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730, , Chapter 9 Infinite Sequences and Series, , 9.1, , Sequences, An idealized superball is dropped from a height of 1 m onto a flat surface. Suppose, that each time the ball hits the surface, it rebounds to two thirds of its previous height., If we let a1 denote the initial height of the ball, a2 denote the maximum height attained, on the first rebound, a3 denote the maximum height attained on the second rebound,, and so on, then we have, 2, a2  ,, 3, , a1  1,, , 4, a3  ,, 9, , a4 , , 8, ,, 27, , p, , (See Figure 1.) This array of numbers, a1, a2, a3, p , is an example of an infinite, sequence, or simply a sequence. If we define the function f by f(x)  1 23 2 x1 and allow, x to take on the positive integral values x  1, 2, 3, p , n, p , then we see that the, sequence a1, a2, a3, p , may be viewed as the functional values of f at these numbers., Thus,, f(1)  1, , 2, 3, , f(2) , , ⏐, ↓, a1, , f(3) , , ⏐, ↓, a2, , 4, 9, , 2 n1, f(n)  a b, 3, , p, , ⏐, ↓, a3, , ⏐, ↓, an, , p, , p, , y (m), 1, 2, 3, 4, 9, 8, 27, , FIGURE 1, The ball rebounds to two, thirds of its previous height, upon hitting the surface., , 0, , 1, , 2, , 3, , 4, , 5, , n, , This discussion motivates the following definition., , DEFINITION Sequence, A sequence {an} is a function whose domain is the set of positive integers. The, functional values a1, a2, a3, p , an, p are the terms of the sequence, and the, term an is called the nth term of the sequence., , Notes, ⬁, 1. The sequence {an} is also denoted by {an}n1, ., 2. Sometimes it is convenient to begin a sequence with ak. In this case the sequence, ⬁, is {an}nk, , and its terms are ak, ak1, ak2, p , an, p ., , EXAMPLE 1 List the terms of the sequence., a. e, , n, f, n1, , b. e, , 1n, 2, , f, n1, , ⬁, c. {(1)n 1n  2}n2, , d. esin, , np ⬁, f, 3 n0

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9.1, , Sequences, , 731, , Solution, a. Here, an  f(n) , a1  f(1) , , n, . Thus,, n1, , 1, 1, 2, 2, 3, 3,  , a2  f(2) ,  , a3  f(3) ,  , p, 11, 2, 21, 3, 31, 4, , and we see that the given sequence can be written as, e, b. e, , 1n, n1, , 2, , fe, , n, 1 2 3 4, n, f  e , , , ,p,, , pf, n1, 2 3 4 5, n1, , 11 12 13 14, 1n, , 1 , 2 , 3 , p , n1 , pf, 0, 2, 2, 2, 2, 2, , ⬁, c. {(1)n 1n  2}n2,  {(1)2 10, (1)3 11, (1)4 12,, , (1) 5 13, p , (1)n 1n  2, p},  {0,  11, 12,  13, p , (1)n 1n  2, p}, Notice that n starts from 2 in this example. (See Note 2 on page 730.), d. esin, , np ⬁, p, 2p, 3p, 4p, 5p, np, f,  esin 0, sin , sin, , sin, , sin, , sin, , p , sin, , pf, 3 n0, 3, 3, 3, 3, 3, 3,  e0,, , 13 13, 13 13, np, ,, , 0, , ,, , p , sin, , pf, 2, 2, 2, 2, 3, , Once again, refer to Note 2., We can often determine the nth term of a sequence by studying the first few terms, of the sequence and recognizing the pattern that emerges., , EXAMPLE 2 Find an expression for the nth term of each sequence., a. e2,, , 1 1 1, b. e1, , , , pf, 8 27 64, , 3, 4, 5, ,, ,, , pf, 12 13 14, , 1 1, 1, c. e1,  , ,  , pf, 2 3, 4, , Solution, a. The terms of the sequence may be written in the form, a1 , , 11, ,, 11, , a2 , , 21, ,, 12, , from which we see that an , , a3 , , 31, ,, 13, , a4 , , 41, ,, 14, , p, , n1, ., 1n, , b. Here,, a1 , so an , , 1, 1, , 3, , ,, , a2 , , 1, 2, , 3, , ,, , a3 , , 1, 3, , 3, , ,, , a4 , , 1, 43, , ,, , p, , 1, , ., n3, c. Note that (1)r is equal to 1 if r is an even integer and 1 if r is an odd integer., Using this result, we obtain, a1 , , (1)0, ,, 1, , a2 , , (1)1, ,, 2, , a3 , , (1)2, ,, 3, , We conclude that the nth term is an  (1)n1>n., , a4 , , (1)3, ,, 4, , p

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732, , Chapter 9 Infinite Sequences and Series, , Some sequences are defined recursively; that is, the sequence is defined by specifying the first term or the first few terms of the sequence and a rule for calculating, any other term of the sequence from the preceding term(s)., , EXAMPLE 3 List the first five terms of the recursively defined sequence a1  2,, a2  4, and an1  2an  an1 for n  2., Solution The first two terms of the sequence are given as a1  2 and a2  4. To find, the third term of the sequence, we put n  2 in the recursion formula to obtain, a3  2a2  a1  2 ⴢ 4  2  6, Next, putting n  3 and n  4 in succession in the recursive formula gives, a4  2a3  a2  2 ⴢ 6  4  8, , and, , a5  2a4  a3  2 ⴢ 8  6  10, , Since a sequence is a function, we can draw its graph. The graphs of the sequences, {n>(n  1)} and {(1)n} are shown in Figure 2. They are just the graphs of the functions f(n)  n>(n  1) and t(n)  (1)n for n  1, 2, 3, p ., g(n), , f(n), , a4, a3, a2 a1, , a2, a4, a6, , y1, , 1, , 1, , 0, 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , a1, a3, a5, , n, , n, (a) The graph of n  1, , {, , }, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , n, , 1, , (b) The graph of {(1)n}, , FIGURE 2, , Limit of a Sequence, If you examine the graph of the sequence {n>(n  1)} sketched in Figure 2a, you will, see that the terms of the sequence seem to get closer and closer to 1 as n gets larger, and larger. In this situation we say that the sequence {n>(n  1)} converges to the, limit 1, written, lim, , n→⬁, , n, 1, n1, , In general, we have the following informal definition of the limit of a sequence., , DEFINITION Limit of a Sequence, A sequence {an} has the limit L, written, lim an  L, , n→⬁, , if an can be made as close to L as we please by taking n sufficiently large. If, lim n→⬁ an exists, we say that the sequence converges. Otherwise, we say that, the sequence diverges.

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9.1, , Sequences, , 733, , A more precise definition of the limit of a sequence follows., , DEFINITION (Precise) Limit of a Sequence, A sequence {an} converges and has the limit L, written, lim an  L, , n→⬁, , if for every e  0 there exists a positive integer N such that 冟 an  L 冟  e whenever n  N., , To illustrate this definition, suppose that a challenger selects an e  0. Then we, must show that there exists a positive integer N such that all points (n, an) on the graph, of {an}, where n  N, lie inside a band of width 2e about the line y  L. (See Figure 3.), y, yL, yL, yL, , L, , FIGURE 3, If n  N, then L  e  an  L  e, or, equivalently, 冟 an  L 冟  e., , 0, , 1 2 3 4 5 6 7 8 9 10 11, (N), , n, , To reconcile this definition with the intuitive definition of a limit, recall that e is, arbitrary. Therefore, by choosing e very small, the challenger ensures that an is “close”, to L. Furthermore, if corresponding to each choice of e, we can produce an N such that, n  N implies that 冟 an  L 冟  e, then we have shown that an can be made as close to, L as we please by taking n sufficiently large., Notice that the definition of the limit of a sequence is very similar to the definition, of the limit of a function at infinity given in Section 3.5. This is expected, since the, only difference between a function f defined by y  f(x) on the interval (0, ⬁) and the, sequence {an} defined by an  f(n) is that n is an integer. (See Figure 4.) This observation tells us that we can often evaluate lim n→⬁ an by evaluating lim x→⬁ f(x), where, f is defined on (0, ⬁) and an  f(n)., y, an, , y  f(x), L, , FIGURE 4, The graph of {an} comprises, the points (n, f(n)) that lie, on the graph of y  f(x) ., , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , 8, , 9 10 11 12 13, , x, n

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734, , Chapter 9 Infinite Sequences and Series, , THEOREM 1, If lim x→⬁ f(x)  L and {an} is a sequence defined by an  f(n), where n is a, positive integer, then lim n→⬁ an  L., You will be asked to prove Theorem 1 in Exercise 75., y, an, , EXAMPLE 4 Find lim, , n→⬁, , 3, , 1, if r  0., nr, , Solution Since an  1>n r, we choose f(x)  1>x r, where x  0. By Theorem 1 in, Section 3.5 we have, , 2, 1, , lim, , x→⬁, , 0, , 1, , 2, , 3, , 4 x, n, , FIGURE 5, The graph of {1>n r} for n  1, 2, 3, 4,, and r  1. The graph of f(x)  1>x r is, shown with a dashed curve., , 1, 0, xr, , Using Theorem 1 of this section, we conclude that, lim, , n→⬁, , 1, 0, nr, , (See Figure 5.), , !, , The converse of Theorem 1 is false. Consider, for example, the sequence, {sin np}  {0}. This sequence evidently converges to 0, since every term of the, sequence is 0. But lim x→⬁ sin px does not exist. (See Figure 6.), y, an, 1, , FIGURE 6, The graph of {sin np} for, n  0, 1, 2, p , 10. The, graph of f(x)  sin px is, shown with a dashed curve., , 0, , 2, , 4, , 6, , 8, , 10 x, n, , 1, , The following limit laws for sequences are the analogs of the limit laws for functions studied in Section 1.2 and are proved in a similar manner., , THEOREM 2 Limit Laws for Sequences, Suppose that lim n→⬁ an  L and lim n→⬁ bn  M and that c is a constant. Then, 1. lim can  cL, n→⬁, , 2. lim (an, n→⬁, , bn)  L, , M, , 3. lim anbn  LM, n→⬁, , 4. lim, , n→⬁, , an, L,  , provided that bn, bn, M, , 0 and M, , 5. lim a pn  Lp, if p  0 and an  0, n→⬁, , 0

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9.1, , Sequences, , 735, , EXAMPLE 5 Determine whether the sequence converges or diverges., a. e, , n, f, n1, , b. {(1)n}, , Solution, a. Both the numerator and the denominator of n>(n  1) approach infinity as n, approaches infinity. So their limits do not exist, and we cannot use Law 4 of, Theorem 2. But we can divide the numerator and denominator by n and then, apply Law 4 to evaluate the resulting limit. Thus,, lim, , n→⬁, , n,  lim, n  1 n→⬁, , 1, 1, 1, n, , 1, , and we conclude that the sequence converges to 1. (See Figure 2a.), b. The terms of the sequence are, 1, 1, 1, 1, p, The sequence evidently does not approach a unique number, and we conclude, that it diverges. (See Figure 2b.), , EXAMPLE 6 Find, a. lim, , n→⬁, , y, an, , ln n, n, , b. lim, , n→⬁, , en, n2, , Solution, a. Observe that both the numerator and the denominator of (ln n)>n approach infinity as n → ⬁ . Therefore, we may not use Law 4 of Theorem 2 directly. Since, an  f(n)  (ln n)>n, we consider the function f(x)  (ln x)>x. Using l’Hôpital’s, Rule, we find, , 0.5, 0.4, 0.3, 0.2, 0.1, 0, , 10, , 20, , lim, , 40 x, n, , 30, , FIGURE 7, The graph of {(ln n)>n} for n  5, 10,, 15, p , 40. The graph of f(x)  (ln x)>x, is shown with a dashed curve., , x→⬁, , 1>x, ln x, 1,  lim,  lim  0, x, x→⬁ 1, x→⬁ x, , Therefore, by Theorem 1 we conclude that, lim, , n→⬁, , (See Figure 7.), b. Once again both en and n 2 approach infinity as n → ⬁ . Choose f(x)  ex>x 2, and, use l’Hôpital’s Rule twice to find that, , y, an, 30, 25, 20, , lim, , x→⬁, , 15, , ex, x2, ,  lim, , x→⬁, , ex, ex,  lim, ⬁, 2x x→⬁ 2, , from which we see that, , 10, 5, 0, , ln n, 0, n, , 1, , 2, , 3, , 4, , 5, , 6, , 7 x, n, , FIGURE 8, The graph of {en>n 2}. The graph of, f(x)  ex>x 2 is shown with a dashed, curve., , lim, , n→⬁, , en, n2, , ⬁, , and we conclude that the sequence {en>n 2} is divergent. (See Figure 8.), The Squeeze Theorem has the following counterpart for sequences. (The proof is, similar to that of the Squeeze Theorem and will be omitted.)

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736, , Chapter 9 Infinite Sequences and Series, , THEOREM 3 Squeeze Theorem for Sequences, If there exists some integer N such that an bn, lim n→⬁ an  lim n→⬁ cn  L, then lim n→⬁ bn  L., , cn for all n  N and, , (See Figure 9.), an, cn, L, , bn, an, , FIGURE 9, The sequence {bn} is squeezed between, the sequences {an} and {cn}., , 0, , EXAMPLE 7 Find lim, , n→⬁, , n, , n!, , where n! (read “n factorial”) is defined by, nn, n!  n(n  1)(n  2) p 1, , Let an  n!>n n. The first few terms of {an} are, , Solution, , a1 , , 1!,  1,, 1, , a2 , , 2!, 2, , 2, , , , 2ⴢ1, ,, 2ⴢ2, , a3 , , 3!, 3, , 3, , , , 3ⴢ2ⴢ1, 3ⴢ3ⴢ3, , and its nth term is, an , , n(n  1) ⴢ p ⴢ 3 ⴢ 2 ⴢ 1, n n1, 3 2 1, n!,  a ba, b ⴢ p ⴢ a ba ba b, n , p, nⴢnⴢ ⴢnⴢnⴢn, n, n, n n n, n, , Therefore,, 0  an, , 1, n, , Since lim n→⬁ 1>n  0, the Squeeze Theorem implies that, lim an  lim, , n→⬁, , n→⬁, , n!, 0, nn, , The next theorem is an immediate consequence of the Squeeze Theorem., , THEOREM 4, If lim n→⬁ 冟 an 冟  0, then lim n→⬁ an  0., , You are asked to prove Theorem 4 in Exercise 76., , 1, n

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9.1, , EXAMPLE 8 Find lim, , n→⬁, , Solution, , Sequences, , 737, , (1)n, ., n, , Since, lim `, , n→⬁, , (1)n, 1, `  lim  0, n, n→⬁ n, , we conclude by Theorem 4 that, lim, , n→⬁, , (1)n, 0, n, , The graph of the sequence {(1)n>n} confirms this result. (See Figure 10.), y, an, 2, , 1, , y  1x, , a2, a4, a5, a3, a1, , FIGURE 10, The terms of the sequence, {(1)n>n} oscillate between the, graphs of y  1>x and y  1>x., , 2, , 4, , 6, , 8, , 10 x, n, , y   1x, , 1, 2, , If we take the composition of a function f with a sequence {an}, we obtain another, sequence {f(an)}. The following theorem shows how to compute the limit of the latter. The proof will be given in Appendix B., , THEOREM 5, If lim n→⬁ an  L and the function f is continuous at L, then, lim f(an)  f( lim an)  f(L), , n→⬁, , Note, , n→⬁, , Compare this theorem with Theorem 4 in Section 1.4., , EXAMPLE 9 Find lim n→⬁ esin(1>n)., Solution, , Observe that esin(1>n)  f(an), where f(x)  ex and an  sin(1>n) . Since, lim sin, , n→⬁, , 1, 0, n, lim sin(1>n), , and f is continuous at 0, Theorem 5 gives lim n→⬁ esin(1>n)  en→⬁, ,  e0  1.

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738, , Chapter 9 Infinite Sequences and Series, , Bounded Monotonic Sequences, Up to now, the convergent sequences that we have dealt with had limits that are readily found. Sometimes, however, we need to show that a sequence is convergent even, if its precise limit is not readily found. Our immediate goal here is to find conditions, that will guarantee that a sequence converges. To do this, we need to make use of two, further properties of sequences., , DEFINITION Monotonic Sequence, A sequence {an} is increasing if, a1  a2  a3  p  an  an1  p, and decreasing if, a1  a2  a3  p  an  an1  p, A sequence is monotonic if it is either increasing or decreasing., , EXAMPLE 10 Show that the sequence e, Solution, , n, f is increasing., n1, , Let an  n>(n  1). We must show that an, , an1 for all n  1; that is,, , n1, (n  1)  1, , n, n1, or, , n1, n2, , n, n1, , To show that this inequality is true, we obtain the following equivalent inequalities:, n(n  2), n  2n, 2, , 0, which is true for n  1. Therefore, an, Alternative Solution, x>(x  1). Since, f ¿(x) , , (n  1)(n  1), , Cross-multiply., , n  2n  1, 2, , 1, an1, so {an} is increasing., , Here, an  f(n)  n>(n  1). So consider the function f(x) , (x  1)(1)  x(1), (x  1), , 2, , , , 1, (x  1)2, , 0, , if, , x0, , we see that f is increasing on (0, ⬁). Therefore, the given sequence is increasing., , EXAMPLE 11 Show that the sequence e n f is decreasing., n, e

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9.1, , Solution, , Sequences, , 739, , We must show that an  an1 for n  1; that is,, n, n1, n , e, en1, nen1  (n  1)en, ne  n  1, , Divide both sides by en., , n(e  1)  1, which is true for all n  1, so {n>en} is decreasing., Next, we explain what is meant by a bounded sequence., , DEFINITION Bounded Sequence, A sequence {an} is bounded above if there exists a number M such that, an, , for all n  1, , M, , A sequence is bounded below if there exists a number m such that, m, , for all n  1, , an, , A sequence is bounded if it is both bounded above and bounded below., , For example, the sequence {n} is bounded below by 0, but it is not bounded above., The sequence {n>(n  1)} is bounded below by 12 and above by 1 and is therefore, bounded. (See Figure 2a.), A bounded sequence need not be convergent. For example, the sequence {(1)n}, is bounded, since 1 (1)n 1; but it is evidently divergent. (See Figure 2b.) Also,, a monotonic sequence need not be convergent. For example, the sequence {n} is increasing and evidently divergent. However, if a sequence is both bounded and monotonic,, then it must be convergent., , THEOREM 6 Monotone Convergence Theorem for Sequences, Every bounded, monotonic sequence is convergent., The plausibility of Theorem 6 is suggested by the sequence {n>(n  1)} whose, graph is reproduced in Figure 11. This sequence is increasing and bounded above by, yn, , a4, a3, a2 a, 1, , FIGURE 11, The increasing, bounded sequence, {n>(n  1)} is convergent., , y1, , 1, , 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , n

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740, , Chapter 9 Infinite Sequences and Series, , any number M  1. Therefore, as n increases, the terms an approach a number (which, is no larger than M) from below. In this case the number is 1, which is also the limit, of this sequence. (A proof of Theorem 6 is given at the end of this section.), Theorem 6 can be used to find the limit of a convergent sequence indirectly, as, the next example shows. It will also play an important role in infinite series (Sections 9.2–9.9)., 2n, n!, , EXAMPLE 12 Show that e f is convergent and find its limit., Solution, a1  2,, , Here, an  2n>n!. The first few terms of the sequence are, a2  2,, , a3 ⬇ 1.333333,, , a6 ⬇ 0.088889,, , a4 ⬇ 0.666667,, , p,, , a5 ⬇ 0.266667,, , a10 ⬇ 0.000282, , These terms suggest that the sequence is decreasing from n  2 onward. To prove this,, we compute, an1, an, , 2n1, (n  1)!, 2n1n!, 2n!, 2, ,  n, , , n, 2, 2 (n  1)!, (n  1)n!, n1, n!, , (1), , So, an1, an, , 1, , if n  1, , Thus, an1 an if n  1, and this proves the assertion. Since all of the terms of the, sequence are positive, {an} is bounded below by 0. Therefore, the sequence is decreasing and bounded below, and Theorem 6 guarantees that it converges to a nonnegative, limit L., To find L, we first use Equation (1) to write, an1 , , 2, an, n1, , (2), , Since lim n→⬁ an  L, we also have lim n→⬁ an1  L. Taking the limit on both sides, of Equation (2) and using Law (3) for limits of sequences, we obtain, L  lim an1  lim a, n→⬁, , n→⬁, , 2, 2, an b  lim, ⴢ lim an  0 ⴢ L  0, n1, n→⬁ n  1 n→⬁, , We conclude that lim n→⬁ 2n>n!  0., Alternative Solution, a2 , , Observe that, , 2ⴢ2, 2ⴢ2ⴢ2, 2, 2ⴢ2ⴢ2ⴢ2, 2 2, 2 2,  2, a3 ,  2a b , a4 ,  a b a b2  2a b ,, 2ⴢ1, 3ⴢ2ⴢ1, 3, 4ⴢ3ⴢ2ⴢ1, 4 3, 3, a5 , , 2ⴢ2ⴢ2ⴢ2ⴢ2, 2 2 2, 2 3,  a b a b a b2  2a b, 5ⴢ4ⴢ3ⴢ2ⴢ1, 5 4 3, 3, , and, an , , 2ⴢ2ⴢ2ⴢpⴢ2, 2 n2, , 2a, b, n ⴢ (n  1) ⴢ (n  2) ⴢ p ⴢ 1, 3

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9.1, , Sequences, , 741, , Therefore, 2 n2, 0  an  2a b, 3, , Since lim n→⬁ 1 23 2 n2  0, the Squeeze Theorem gives the desired result., The next example contains some important results that we will derive here using, the Squeeze Theorem. (You will also be asked to demonstrate their validity using the, properties of exponential functions in Exercise 77.), , EXAMPLE 13 Show that lim n→⬁ r n  0 if 冟 r 冟  1., Solution If r  0, then each term of the sequence {r n} is 0, and the sequence converges to 0. Now suppose that 0  冟 r 冟  1. Then 1> 冟 r 冟 is greater than 1. So there exists, a positive number p such that, 1, 1p, 冟r冟, Using the Binomial Theorem, we have, (1  p)n  1  np , , n(n  1) 2 p, p   p n  np, 2!, , Thus,, 0  冟 r 冟n , , 1, 1, , np, (1  p)n, , But, lim, , n→⬁, , 1, 0, np, , so by the Squeeze Theorem, lim 冟 r 冟n  0, , n→⬁, , Finally, using Theorem 4, we conclude that lim n→⬁ r n  0., If r  1, then r n  1 for all n, and the sequence {r n} evidently converges to 1. If, r  1, then the sequence {r n}  {(1)n} is divergent. (See Example 5b.) If 冟 r 冟  1,, then 冟 r 冟  1  p for some positive number p. Using the Binomial Theorem again, we, have, 冟 r 冟n  (1  p)n  np, Since p  0, lim n→⬁ np  ⬁ . This shows that {r n} diverges if 冟 r 冟  1., A summary of these results follows., Properties of the Sequence {r n}, The sequence {r n} converges if 1  r, lim r n  e, , n→⬁, , It diverges for all other values of r., , 1 and, , 0 if 1  r  1, 1 if r  1

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742, , Chapter 9 Infinite Sequences and Series, , Proof of Theorem 6, The proof of Theorem 6 depends on the Completeness Axiom for the real number system, which states that every nonempty set S of real numbers that is bounded above has, a least upper bound. Thus, if x M for all x in S, then there must be a real number b, such that b is an upper bound of S (x b for all x 僆 S), and if N is any upper bound, of S, then N  b. For example, if S is the interval (2, 3), then the number 4 (or any, number greater than 3) is an upper bound of S and 3 is the least upper bound of S. As, a consequence of this axiom, it can be shown that every nonempty set of real numbers, that is bounded below has a greatest lower bound. The Completeness Axiom merely, states that the real number line has no gaps or holes., , PROOF Suppose that {an} is an increasing sequence. Since {an} is bounded, the set, S  {an 冟 n  1} is bounded above, and by the Completeness Axiom it has a least upper, bound L. (See Figure 12.) We claim that L is the limit of the sequence., an, L, an, L, , a3, a2, a1, 0, , FIGURE 12, An increasing sequence bounded above, must converge to its least upper bound., , 1, , 2, , 3, N, , N1, N2, , n, , To show this, let e  0 be given. Then L  e is not an upper bound of S (since, L is the least upper bound of S). Therefore, there exists an integer N such that, aN  L  e. But the sequence is increasing, so an  aN for every n  N. In other, words, if n  N, we have an  L  e. Since an L,, 0, , L  an  e, , This shows that, 冟 L  an 冟  e, whenever n  N, so lim n→⬁ an  L., The proof is similar for the case in which {an} is decreasing, except that we use, the greatest lower bound instead., , 9.1, , CONCEPT QUESTIONS, , 1. Explain each of the following terms in your own words, and, give an example of each., a. Sequence, b. Convergent sequence, c. Divergent sequence, d. Limit of a sequence, , 2. Explain each of the following terms in your own words, and, give an example of each., a. Bounded sequence, b. Monotonic sequence

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9.1, , 9.1, , 33. an , , 2n, 3 1, , 34. an , , 2n  1, en, , n, , n1, 2n  1, , 2. an , , (1)n12n, n1, , 35. an  1n  1  1n, , 36. an , , np, ,, en, , np, 2, , 4. an , , 1 ⴢ 3 ⴢ 5 ⴢ p ⴢ (2n  1), n!, , 2 1>n, 37. an  a1  b, n, , 38. an , , (2)n, n!, , 3. an  sin, , 2n, (2n)!, , 39. an , , sin2 n, 1n, , In Exercises 7–12, find an expression for the nth term of the, sequence. (Assume that the pattern continues.), , 40. an , , 1 ⴢ 3 ⴢ 5 ⴢ p ⴢ (2n  1), n!, , 3 4 5 6 7, 8. e , , , , , pf, 4 9 16 25 36, , 41. an , , 5. an , , 6. a1  2,, , 1 2 3 4 5, 7. e , , , , , pf, 2 3 4 5 6, , an1  3an  1, , 1, 1 1, 1, , pf, 9. e1, ,  , , , 2, 6 24, 120, 10. {0, 2, 0, 2, 0, p}, 1, 2, 3, 4, 5, ,, ,, ,, ,, ,pf, 11. e, 1ⴢ2 2ⴢ3 3ⴢ4 4ⴢ5 5ⴢ6, 1 1ⴢ3 1ⴢ3ⴢ5 1ⴢ3ⴢ5ⴢ7 1ⴢ3ⴢ5ⴢ7ⴢ9, ,, ,, ,, , pf, 12. e ,, 2 2 ⴢ 4 2 ⴢ 4 ⴢ 6 2 ⴢ 4 ⴢ 6 ⴢ 8 2 ⴢ 4 ⴢ 6 ⴢ 8 ⴢ 10, In Exercises 13–42, determine whether the sequence {an} converges or diverges. If it converges, find its limit., 2n, 13. an , n1, 15. an  1  2(1)n, n1, 2n  1, , 17. an , n, n2, 19. an , , 2n 2  3n  4, 3n  1, 2, , 14. an  1n  1, (1), , 16. an  1 , 18. an , , n, , n 3>2, n2  1, , 20. an  (1)n, 22. an , , 23. an , , 2n, 1n  1, , 2 n, 24. an  1  a b, e, , n  2n 2>3, , 27. an  sin, 29. an , , np, 2, , sin 1n, 1n, , 31. an  tanh n, , 2, , 22n 2  1, n, , n2, , , , 3, , n2, , In Exercises 43–48, (a) graph the sequence {an} with a graphing, utility, (b) use your graph to guess at the convergence or divergence of the sequence, and (c) use the properties of limits to, verify your guess and to find the limit of the sequence if it, converges., 43. an , , n1, n2, , 44. an  (1)n, , 45. an , , n!, nn, , 46. an  2 tan1 a, , 47. an  n sin, , 30. an  tan1 n 2, 32. an , , ln n 2, 1n, , n1, b, n3, , 49. Evaluate, , lim, , n→⬁, , 1 9, 1  a1  b, n, 1, 1  a1  b, n, , Hint: Use Theorem 1., , 50. Evaluate, lim na1 , , 26. an  cos np  2, np, b, 2n  1, , 2n  1, n3, , 2 n, 48. an  a1  b, n, , 1, n, , n→⬁, , 28. an  sina, , p0, , n, p 2, n, 123pn, n, , 42. an , n2, 2, , , n2, 3n  1, , 2  (1)n, n, , n 1>2  n 1>3, , 1, , n2, , 2n 2  1, , 21. an , , 25. an , , 743, , EXERCISES, , In Exercises 1–6, write the first five terms of the sequence {an}, whose nth term is given., 1. an , , Sequences, , 7, , B, , 1, 1 b, n, , Hint: Use Theorem 1., , In Exercises 51–58, determine whether the sequence {an} is, monotonic. Is the sequence bounded?, 51. an , , 3, 2n  5, , 53. an  3 , , V Videos for selected exercises are available online at www.academic.cengage.com/login., , 1, n, , 52. an , , 2n, n1, , 54. an  2 , , (1)n, n

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744, , Chapter 9 Infinite Sequences and Series, , 55. an , , sin n, n, , 56. an  tan1 n, , 57. an , , n, 2n, , 58. an , , ln n, n, , 59. Compound Interest If a principal of P dollars is invested in, an account earning interest at the rate of r per year compounded monthly, then the accumulated amount An at the, end of n months is, r n, An  Pa1  b, 12, a. Write the first six terms of the sequence {An} if, P  10,000 and r  0.105. Interpret your results., b. Does the sequence {An} converge or diverge?, 60. Quality Control Half a percent of the microprocessors manufactured by Alpha Corporation for use in regulating fuel, consumption in automobiles are defective. It can be shown, that the probability of finding at least one defective microprocessor in a random sample of n microprocessors is, f(n)  1  (0.995)n. Consider the sequence {an} defined by, an  f(n) ., a. Write the terms a10, a100, and a1000., b. Evaluate lim n→⬁ an, and interpret your result., 61. Annuities An annuity is a sequence of payments made at, regular intervals. Suppose that a sum of $200 is deposited, at the end of each month into an account earning interest, at the rate of 12% per year compounded monthly. Then the, amount on deposit (called the future value of the annuity) at, the end of the nth month is f(n)  20,000[(1.01) n  1]., Consider the sequence {an} defined by an  f(n)., a. Find the 24th term of the sequence {an}, and interpret, your result., b. Evaluate lim n→⬁ an, and interpret your result., 62. Continuously Compounded Interest If P dollars is invested in an, account paying interest at the rate of r per year compounded, m times per year, then the accumulated amount at the end, of t years is, Am  Pa1 , , r mt, b, m, , m  1, 2, 3, p, , a. Find the limit of the sequence {Am}., b. Interpret the result in part (a)., Note: In this situation, interest is said to be compounded, , 64. Define the sequence {an} recursively by a0  2 and, an1  1an for n  1., n, a. Show that an  21>2 ., b. Evaluate lim n→⬁ an., c. Verify the result of part (b) graphically., 65. Newton’s Method Suppose that A  0. Applying Newton’s, method to the solution of the equation x 2  A  0 leads to, the sequence {x n} defined by, x n1 , , 1, A, ax n  b, xn, 2, , a. Show that if L  lim n→⬁ x n exists, then L  1A., Hint: lim x n1  L, n→⬁, , b. Find 15 accurate to four decimal places., 66. Finding the Roots of an Equation Suppose that we want to find a, root of f(x)  0. Newton’s method provides one way of, finding it. Here is another method that works under suitable, conditions., a. Write f(x)  0 in the form x  t(x), where t is continuous. Then generate the sequence {x n} by the recursive, formula x n1  t(x n), where x 0 is arbitrary., b. Show that if the sequence {x n} converges to a number r,, then r is a solution of f(x)  0., Hint: lim x n1  r, n→⬁, , c. Use this method to find the root of f(x)  3x 3  9x  2, (accurate to four decimal places) that lies in the interval, (0, 1)., Hint: Write 3x 3  9x  2  0 in the form x  19 (3x 3  2). Take, x 0  0., , 67. A Floating Object A sphere of radius 1 ft is made of wood that, has a specific gravity of 23. If the sphere is placed in water, it, sinks to a depth of h ft. It can be shown that h satisfies the, equation, h3  3h2 , , Hint: Show that the equation can be written in the form, h  13 23h3  8., , 68. Find the limit of the sequence, , 5 12, 22 12, 322212, p 6, n1)>2n, , c. What is the accumulated amount at the end of 3 years if, $1000 is invested in an account paying interest at the rate, of 10% per year compounded continuously?, 2 3n, 63. Find the limit of the sequence e a1  b f . Confirm your, n, results visually by plotting the graph of, 2 3x, f(x)  a1  b, x, , 8, 0, 3, , Use the method described in Exercise 66 to find h accurate, to three decimal places., , Hint: Show that an  2(2, , continuously., , x0  0, , n, ,  211>2 ., , 69. Consider the sequence {an} defined by a1  12 and, an  12  an1 for n  2. Assuming that the sequence, converges, find its limit., Note: Using the principle of mathematical induction, it can be, shown that {an} is increasing and bounded by 2 and, hence, by, Theorem 6 is convergent.

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9.1, 70. Show that if lim n→⬁ a2n  L and lim n→⬁ a2n1  L, then, lim n→⬁ an  L., 71. Let the sequence {an} be defined by, an  1 , , 1, 2, , 2, , , , Hint:, , 1, n2, , , , 1, p 2, 3, n, , lim an , , n→⬁, , 2, , 1, 1, 1, ,  , for n  2, n(n  1), n1, n, , c. Using the results of parts (a) and (b), what can you, deduce about the convergence of {an}?, , 745, , 78. Fibonacci Sequence The Fibonacci sequence {Fn} is defined, by F1  1, F2  1, and Fn1  Fn  Fn1 for n  2. Let, an  Fn1>Fn. Assuming that {an} is convergent, show that, , 1, , a. Show that {an} is increasing., b. Show that {an} is bounded above by establishing that, an  2  1>n for n  2., , Sequences, , 1, (1  15), 2, , Hint: First, show that an1  1  1>an2. Then use the fact that if, lim n→⬁ an  L, then lim n→⬁ an2  lim n→⬁ an1  L., , Note: The number 12 (1  15), which is approximately 1.6, has the, following special property: A picture with a ratio of width to height, equal to this number is especially pleasing to the eye. The ancient, Greeks used this “golden” ratio in designing their beautiful temples, and public buildings, such as the Parthenon., , 72. Let the sequence {an} be defined by, 1, 1, 1, p n,  2, 21, 2 n, 2 2, , a. Show that {an} is increasing., b. Show that {an} is bounded above., c. Using the results of parts (a) and (b), what can you, deduce about the convergence of {an}?, 73. Let the sequence {an} be defined by, a1 , , a0, ,, 2  a0, , a2, a3 , ,, 2  a2, , p,, , a2 , , a1, ,, 2  a1, , an1, an , ,, 2  an1, , The front of the Parthenon has a ratio of width to, height that is approximately 1.6 to 1., p, , where an  0., a. Show that {an} is convergent., b. Find the limit of {an}., lim 1a  1, , n→⬁, , In Exercises 79–86, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, explain why, or give an example to show why it is false., 79. If {an} and {bn} are divergent, then {an  bn} is divergent., , 74. Use the Squeeze Theorem for Sequences to prove that, n, , The Gallery Collection/Corbis, , an , , a0, , Hint: For n sufficiently large, 1>n  a  n., , 75. Prove Theorem 1: If lim x→⬁ f(x)  L and {an} is a sequence, defined by an  f(n), where n is a positive integer, then, lim n→⬁ an  L., 76. Prove Theorem 4: If lim n→⬁ 冟 an 冟  0, then lim n→⬁ an  0., 77. Prove the properties of the sequence {r n} given on, page 741 using the results lim x→⬁ a x  0 if 0  a  1, and lim x→⬁ a x  ⬁ if a  1., , 80. If {an} is divergent, then {冟 an 冟} is divergent., 81. If {an} converges to L and {bn} converges to 0, then {anbn}, converges to 0., , 82. If {an} converges and {bn} converges, then {an>bn} converges., 83. If {an} is bounded and {bn} converges, then {anbn} converges., 84. If {an} is bounded, then {an>n} converges to 0., , 85. If lim n→⬁ anbn exists, then both lim n→⬁ an and lim n→⬁ bn, exist., 86. If lim n→⬁ 冟 an 冟 exists, then lim n→⬁ an exists.

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746, , Chapter 9 Infinite Sequences and Series, , 9.2, , Series, Consider again the example involving the bouncing ball. Earlier we found a sequence, describing the maximum height attained by the ball on each rebound after hitting a surface. The question that follows naturally is: How do we find the total distance traveled, by the ball? To answer this question, recall that the initial height and the heights attained, on each subsequent rebound are, 1,, , 2 2, a b ,, 3, , 2, ,, 3, , 2 3, a b , p, 3, , meters, respectively. (See Figure 1.) Observe that the distance traveled by the ball when, it first hits the surface is 1 m. When it hits the surface the second time, it will have, traveled a total distance of, 2, 1  2a b, 3, , 1, , or, , 4, 3, , meters. When it hits the surface the third time, it will have traveled a distance of, 2, 2 2, 1  2a b  2a b, 3, 3, , or, , 1, , 4, 8, , 3, 9, , meters. Continuing in this fashion, we see that the total distance traveled by the ball is, 2, 2 2, 2 3, 1  2a b  2a b  2a b  p, 3, 3, 3, , (1), , meters. Observe that this last expression involves the sum of infinitely many terms., y (m), 1, 2, 3, 4, 9, 8, 27, , 0, , FIGURE 1, , 1, , 2, , 3, , 4, , 5, , n, , In general, an expression of the form, a1  a2  a3  p  an  p, is called an infinite series or, more simply, a series. The numbers a1, a2, a3, p are, called the terms of the series; an is called the nth term, or general term, of the series;, and the series itself is denoted by the symbol, ⬁, , a an, n1, , or simply 兺 an.

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9.2, , Series, , 747, , How do we define the “sum” of an infinite series, if it exists? To answer this question, we use the same technique that we have employed several times before: using, quantities that we can compute to help us define new ones. For example, in defining, the slope of the tangent line to the graph of a function, we take the limit of the slope, of secant lines (quantities that we can compute); and in defining the area under the, graph of a function, we take the limit of the sum of the area of rectangles (again, quantities that we can compute). Here, we define the sum of an infinite series as the limit, of a sequence of finite sums (quantities that we can compute)., We can get an inkling of how this may be done from examining the series (1) giving the total distance traveled by the ball. Define the sequence {Sn} by, S1  1, 2, S2  1  2a b, 3, 2, 2 2, S3  1  2a b  2a b, 3, 3, o, 2, 2 2, 2 n1, Sn  1  2a b  2a b  p  2a b, 3, 3, 3, giving the total vertical distance traveled by the ball when it hits the surface the first, time, the second time, the third time, p , and the nth time, respectively. If the series, (1) has a sum S (the total distance traveled by the ball), then the terms of the sequence, {Sn} form a sequence of increasingly accurate approximations to S. This suggests that, we define, S  lim Sn, n→⬁, , We will complete the solution to this problem in Example 5., Motivated by this discussion, we define the sum of an infinite series., , DEFINITION Convergence of Infinite Series, Given an infinite series, ⬁, , p, p, a an  a1  a2  a3   an , n1, , the nth partial sum of the series is, n, , Sn  a ak  a1  a2  a3  p  an, k1, , If the sequence of partial sums {Sn} converges to the number S, that is, if, lim n→⬁ Sn  S, then the series 兺 an converges and has sum S, written, ⬁, , p, p, a an  a1  a2  a3   an   S, n1, , If {Sn} diverges, then the series 兺 an diverges.

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748, , Chapter 9 Infinite Sequences and Series, , !, , Be sure to note the difference between a sequence and a series. A sequence is a, succession of terms, whereas a series is a sum of terms., , EXAMPLE 1 Determine whether the series converges. If the series converges, find, its sum., ⬁, , a. a n, n1, , ⬁, 1, 1, b. a a , b, n1, n1 n, , Solution, a. The nth partial sum of the series is, Sn  1  2  3  p  n , , n(n  1), 2, , Since, lim Sn  lim, , n→⬁, , n→⬁, , n(n  1), ⬁, 2, , ⬁, we conclude that the limit does not exist and 兺n1, n diverges., b. The nth partial sum of the series is, , 1, 1, 1, 1, 1, 1, 1, 1, 1, Sn  a1  b  a  b  a  b  p  a,  ba , b, n, n, 2, 2, 3, 3, 4, n1, n1, Removing the parentheses, we see that all the terms of Sn, except for the first and, last, cancel out. So, Sn  1 , , 1, n1, , Since, lim Sn  lim a1 , , n→⬁, , n→⬁, , 1, b1, n1, , we conclude that the series converges and has sum 1, that is,, ⬁, , 1, 1, a an  n  1b  1, n1, The series in Example 1b is called a telescoping series., ⬁, , 4, , EXAMPLE 2 Show that the series a 2, is convergent, and find its sum., n1 4n  1, Solution First, we use partial fraction decomposition to rewrite the general term, an  4>(4n 2  1):, an , , 4, 4n  1, 2, , , , 4, 2, 2, , , (2n  1)(2n  1), 2n  1, 2n  1

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9.2, , Series, , 749, , Then we write the nth partial sum of the series as, n, , Sn  a, k1, , n, 2, 2, , a a 2k  1  2k  1 b, 2, 4k  1, k1, , 4, , 2, 2, 2, 2, 2, 2, 2, 2, a  ba  ba  bpa, , b, 1, 3, 3, 5, 5, 7, 2n  1, 2n  1, 2, , 2, 2n  1, , This is a telescoping series., , Since, lim Sn  lim a2 , , n→⬁, , n→⬁, , 2, b2, 2n  1, , we conclude that the given series is convergent and has sum 2; that is,, ⬁, , a, n1, , 4, 4n  1, 2, , 2, , Geometric Series, Geometric series play an important role in mathematical analysis. They also arise frequently in the field of finance. The convergence or divergence of a geometric series is, easily established., , DEFINITION Geometric Series, A series of the form, ⬁, , a ar, , n1, ,  a  ar  ar 2  p  ar n1  p, , a, , 0, , n1, , is called a geometric series with common ratio r., The following theorem tells us the conditions under which a geometric series is, convergent., , THEOREM 1, If 冟 r 冟  1, then the geometric series, ⬁, , a ar, , n1, ,  a  ar  ar 2  p  ar n1  p, , n1, ⬁, a, converges, and its sum is a ar n1 , . The series diverges if 冟 r 冟  1., 1r, n1, , PROOF The nth partial sum of the series is, Sn  a  ar  ar 2  p  ar n1, Multiplying both sides of this equation by r gives, rSn  ar  ar 2  ar 3  p  ar n

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750, , Chapter 9 Infinite Sequences and Series, , Subtracting the second equation from the first then yields, (1  r)Sn  a  ar n  a(1  r n), If r, , 1, we can solve the last equation for Sn, obtaining, Sn , , a(1  r n), 1r, , From Example 13 on page 741 we know that lim r n  0 if 冟 r 冟  1, so, n→⬁, , lim Sn  lim, , n→⬁, , n→⬁, , a(1  r n), a, , 1r, 1r, , This implies that, ⬁, , a ar, , n1, , n1, , , , a, 1r, , 冟r冟  1, , If 冟 r 冟  1, then the sequence {r n} diverges, so lim n→⬁ Sn does not exist. This means, that the geometric series diverges. We leave it as an exercise to show that {Sn} diverges, if r  1, so the series also diverges for these values of r., , EXAMPLE 3 Determine whether the series converges or diverges. If it converges, find, its sum., ⬁, 1 n1, 3, 3, 3, a. a 3a b, 3   p, 2, 2, 4, 8, n1, ⬁, 4 n1, 20, 80, 320 p, b. a 5a b, 5, , , , 3, 3, 9, 27, n1, , Solution, a. This is a geometric series with a  3 and common ratio r  12. Since 冟 12 冟  1,, Theorem 1 tells us that the series converges and has sum, ⬁, , 1 n1, 3, 3a, b, , 2, a, 2, 1  1 12 2, n1, , The graphs of {an} and {Sn} for this series are shown in Figure 2a., b. This is a geometric series with a  5 and common ratio r  43. Since 43  1,, Theorem 1 tells us that the series is divergent. The graphs of {an} and {Sn} for, this series are shown in Figure 2b., 3, , 2000, , {Sn}, 1000, 0, , {an}, , 1.5, , FIGURE 2, , (a) The geometric series converges, because | r |  1., , {Sn}, , 20.5, {an}, 0, (b) The geometric series diverges, because | r |  1., , 20.5

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9.2, , Series, , 751, , EXAMPLE 4 Express the number 3.214  3.2141414 p as a rational number., Solution, , We rewrite the number as, 3.2141414 p  3.2 , , 14, 3, , 10, , , , 14, 5, , 10, , , , 14, 107, , p, , , , 32, 14, 1, 1,  3 c1  2  4  p d, 10, 10, 10, 10, , , , ⬁, 32, 14, 1 n1,  a a 3b a 2b, 10, 10, n1 10, , 14, 1, The expression after the first term is a geometric series with a  1000, and r  100, . Using, Theorem 1, we have, 14, , 3.2141414 p , , , 32, 1000, , 1, 10, 1  100, 32, 14, 3182, , , 10, 990, 990, , EXAMPLE 5 Complete the solution of the bouncing ball problem that was introduced, at the beginning of this section. Recall that the total vertical distance traveled by the, ball is given by, 2, 2 2, 2 3, 1  2a b  2a b  2a b  p, 3, 3, 3, meters., Solution, , If we let d denote the total vertical distance traveled by the ball, then, ⬁, 4 2 n1, d  1  a a ba b, 3, n1 3, , The expression after the first term is a geometric series with a  43 and r  23. Using, Theorem 1, we obtain, d1, , 4, 3, , 1  23, , 145, , and conclude that the total distance traveled by the ball is 5 m., , The Harmonic Series, The series, ⬁, 1, 1, 1 p, 1, a n1234, n1, , is called the harmonic series. Before showing that this series is divergent, we make, this observation: If a sequence {bn} is convergent, then any subsequence obtained by, deleting any number of terms from the parent sequence {bn} must also converge to the, same limit. Therefore, to show that a sequence is divergent, it suffices to produce a, subsequence of the parent sequence that is divergent.

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752, , Chapter 9 Infinite Sequences and Series, , In keeping with this strategy, let us show that the subsequence, S2, S4, S8, S16, p , S2n, p, of the sequence {Sn} of partial sums of the harmonic series is divergent. We have, S2  1 , , 1, 2, , S4  1 , , 1, 1, 1, 1, 1, 1, 1,  a  b  1   a  b  1  2a b, 2, 3, 4, 2, 4, 4, 2, , S8  1 , , 1, 1, 1, 1, 1, 1, 1, a  ba    b, 2, 3, 4, 5, 6, 7, 8, , 1, S16  1 , 1, , 1, 1, 1, 1, 1, 1, 1, a  ba p ba p b, 2, 3, 4, 5, 8, 9, 16, 1, 1, 1, 1, 1, 1, 1, a  ba p ba p b, 2, 4, 4, 8, 8, 16, 16, ⎫, ⎪, ⎪, ⎬, ⎪, ⎪, ⎭, , ⎫, ⎪, ⎪, ⎪, ⎬, ⎪, ⎪, ⎪, ⎭, , 1, , 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,  a  b  a    b  1     1  3a b, 2, 4, 4, 8, 8, 8, 8, 2, 2, 2, 2, , 4 terms, , 8 terms, , 1, 1, 1, 1, 1,     1  4a b, 2, 2, 2, 2, 2, , and, in general, S2n  1  n 1 12 2 . Therefore,, , lim S2n  ⬁, , n→⬁, , so {Sn} is divergent. This proves that the harmonic series is divergent., , The Divergence Test, The next theorem tells us that the terms of a convergent series must ultimately approach, zero., , THEOREM 2, ⬁, If 兺n1, an converges, then lim n→⬁ an  0., , PROOF We have Sn  a1  a2  p  an1  an  Sn1  an, so an  Sn  Sn1., ⬁, Since 兺n1, an is convergent, the sequence {Sn} is convergent. Let lim n→⬁ Sn  S. Then, , lim an  lim (Sn  Sn1)  lim Sn  lim Sn1  S  S  0, , n→⬁, , n→⬁, , n→⬁, , n→⬁, , The Divergence Test is an important consequence of Theorem 2., , THEOREM 3 The Divergence Test, If lim n→⬁ an does not exist or lim n→⬁ an, , ⬁, 0, then 兺n1, an diverges.

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9.2, , 753, , Series, , ⬁, The Divergence Test does not say that if lim n→⬁ an  0, then 兺n1, an must converge. In other words, the converse of Theorem 2 is not true in general. For example,, ⬁, lim n→⬁ 1>n  0, yet the harmonic series 兺n1, 1>n is divergent. In short, the Divergence Test rules out convergence for a series whose nth term does not approach zero, but yields no information if an does approach zero—that is, the series might or might, not converge., , EXAMPLE 6 Show that the following series are divergent., ⬁, , ⬁, , a. a (1)n1, , b. a, , n1, , n1, , 2n 2  1, 3n 2  1, , Solution, a. Here, an  (1)n1, and, lim an  lim (1)n1, , n→⬁, , n→⬁, , does not exist. We conclude by the Divergence Test that the series diverges., 2n 2  1, b. Here, an  2, , and, 3n  1, , lim an  lim, , n→⬁, , n→⬁, , 2n 2  1, 3n  1, 2, , 2,  lim, , n→⬁, , 3, , 1, 2, n2, , 1, 3, , 0, , n2, , so by the Divergence Test, the series diverges., , Properties of Convergent Series, The following properties of series are immediate consequences of the corresponding, properties of the limits of sequences. We omit the proofs., , THEOREM 4 Properties of Convergent Series, ⬁, ⬁, If 兺n1, an  A and 兺n1, bn  B are convergent and c is any real number, then, ⬁, ⬁, 兺n1 can and 兺n1(an bn) are also convergent, and, ⬁, , ⬁, , a. a can  c a an  cA, n1, , ⬁, , ⬁, , b. a (an, , n1, , n1, , ⬁, , bn)  a an, n1, , ⬁, , a bn  A, , B, , n1, , EXAMPLE 7 Show that the series a c,  n d is convergent, and find its, 3, n1 n(n  1), sum., 2, , 4, , ⬁, Solution First, consider the series 兺n1, 1>[n(n  1)]. Using partial fraction decomposition, we can write this series in the form, ⬁, , ⬁, 1, 1, 1, , a n(n  1), a an  n  1b, n1, n1

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754, , Chapter 9 Infinite Sequences and Series, , Using the result of Example 1, we see that, ⬁, , 1, a n(n  1)  1, n1, ⬁, 4, Next, observe that a n is a geometric series with a  43 and r  13, so, n1 3, ⬁, , 4, , 4, 3, 2, a 3n , 1  13, n1, Therefore, by Theorem 4 the given series is convergent, and, ⬁, ⬁, ⬁, 2, 4, 1, 4, a c n(n  1)  3n d  2 a n(n  1)  a 3n, n1, n1, n1, , 2ⴢ120, , 9.2, , CONCEPT QUESTIONS, ⬁, 2. Suppose that 兺n1, an  6., a. Evaluate lim n→⬁ Sn, where Sn is the nth partial sum of, ⬁, 兺n1, an., ⬁, b. Find 兺n2, an if it is known that a1  12., , 1. Explain the difference between, a. A sequence and a series, b. A convergent sequence and a convergent series, c. A divergent sequence and a divergent series, d. The limit of a sequence and the sum of a series, , 9.2, , EXERCISES, , In Exercises 1–6, find the nth partial sum Sn of the telescoping, series, and use it to determine whether the series converges or, diverges. If it converges, find its sum., ⬁, 1, 1, 1. a a,  b, n, n2 n  1, , ⬁, 1, 1, 2. a a, , b, 2n  1, n1 2n  3, , ⬁, 4, 3. a, n1 (2n  3)(2n  5), , ⬁, 8, b, 4. a a 2, n1 4n  4n  3, , ⬁, 1, 1, , b, 5. a a, ln(n  1), n2 ln n, , ⬁, 2, 6. a, n1 1n  1  1n, , In Exercises 7–14, determine whether the geometric series, converges or diverges. If it converges, find its sum., 7. 4 , 9., , 8, 16, 32 p, , , , 3, 9, 27, , 5, 5, 5, 5,  , , p, 3, 9, 27, 81, ⬁, , 1 n, b, 11. a 2a, 12, n0, ⬁, , 13. a 2n3n1, n0, , 8. , , 1, 1, 1, 1,   , p, 2, 4, 8, 16, , 10. 1 , ⬁, , 4, 16, 64 p, , , , 3, 9, 27, en, , 12. a n1, n1 3, ⬁, , 14. a (1)n13n21n, , In Exercises 15–22, show that the series diverges., 15., , 1, 2, 3,   p, 2, 3, 4, , ⬁, 2n, 17. a, 3n, 1, n1, ⬁, , 19. a 2(1.5) n, n1, , 16. 1 , , 3, 9, 27 p,  , , 2, 4, 8, , ⬁, n2, 18. a, 2, n1 2n  1, ⬁, (1)n3n, 20. a, 2n1, n0, , ⬁, , 1, 21. a, 2, , 3n, n1, , ⬁, , 22. a, n1, , n, 22n 2  1, , In Exercises 23–28, (a) compute as many terms of the sequence, of partial sums, Sn , as is necessary to convince yourself that, the series converges or diverges. If it converges, estimate its sum., (b) Plot {Sn} to give a visual confirmation of your observation, in part (a). (c). If the series converges, find the exact sum. If it, diverges, prove it, using the Divergence Theorem., ⬁, 6, 23. a, n(n,  1), n1, ⬁, , 7, 25. a 3a b, 8, , n1, , n1, , n1, , ⬁, , 27. a sin n 2, n1, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , ⬁, , 24. a, , 2n, , 2n 2  1, 2 n1, 26. a 5a b, 3, n1, n1, ⬁, , ⬁, 1, 1, 28. a a n  n b, 3, n1 2

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9.2, In Exercises 29–54, determine whether the given series, converges or diverges. If it converges, find its sum., ⬁, 1, 29. a, n(n,  2), n1, , ⬁, 1, 30. a 2, n2 n  1, , ⬁, 2n, 31. a n, n0 5, , 32. a, , ⬁, , ⬁, n0, , (3)n, , 33. a n1, n0 2, , 3n1, 5n, , ⬁, , n1, ⬁, , 36. a, n0, , ⬁, , ⬁, , 37. a 3(1.01)n, , 38. a, , 3n 2  2, 3n  1, , n1, , 3n1, , ⬁, , ⬁, , n, , ⬁, , 2, 41. a, n, n1 1  (0.2), , 65. Probability of Winning a Dice Toss Peter and Paul take turns tossing a pair of dice. The first person to throw a 7 wins. If, Peter starts the game, then it can be shown that his chances, of winning are, , 2n 2  n  1, , n0, , 1, 1, d, 39. a c n , n(n  1), n1 2, , (1) 2, 2, d, 40. a c n1 , 3n1, n1 3, ⬁, , n, b, 42. a lna, n1, n1, , ⬁, , 45. a [2(0.1) n  3(1)n(0.2)n], n1, , ⬁, , 3 n, e n1, 46. a c a b  a b d, p, 3, n0, ⬁, , 2n  5n, b, 48. a a, 3n, n1, ⬁, , 50. a sin2 n, n1, , ⬁, , 2n  3n, b, 47. a a, 6n, n0, ⬁, , 49. a tan1 n, n1, ⬁, 1, 51. a n sin, n, n1, , ⬁, , sin n, 52. a, 1,  en, n1, , ⬁, n, 53. a, ln, n, n2, , ⬁, 2 n, 54. a a1  b, n, n1, , 55. 0.4  0.444 p, , 56. 0.23  0.232323 p, , 57. 1.213  1.213213213 p, , 58. 3.14234  3.142343434 p, , In Exercises 59–62, find the values of x for which the series converges, and find the sum of the series. (Hint: First show that the, series is a geometric series.), 59. a (x), , ⬁, , n, , 60. a (x  2), , n0, , n0, , ⬁, , ⬁, x 2n, 62. a n, n0 3, , 61. a 2n (x  1)n, n1, , n, , 63. Distance Traveled by a Bouncing Ball A rubber ball is dropped, from a height of 2 m onto a flat surface. Each time the ball, hits the surface, it rebounds to half its previous height. Find, the total distance the ball travels., , 1, 1 5 2, 1 5 4,  a ba b  a ba b  p, 6, 6 6, 6 6, , Find p., 66. Multiplier Effect of a Tax Cut Suppose that the average wage, earner saves 9% of his or her take-home pay and spends the, other 91%. What is the estimated impact that a proposed, $30 billion tax cut will have on the economy over the long, run because of the additional spending generated by the proposed tax cut?, Note: This phenomenon in economics is known as the multiplier, effect., , 67. Perpetuities An annuity is a sequence of payments that are, made at regular time intervals. If the payments are allowed, to continue indefinitely, then it is a perpetuity., a. Suppose that P dollars is paid into an account at the, beginning of each month and that the account earns, interest at the rate of r per year compounded monthly., Then the present value V of the perpetuity (that is, the, value of the perpetuity in today’s dollars) is, V  Pa1 , , In Exercises 55–58, express each number as a rational number., , ⬁, , p, , n1 n, , ⬁, 1, 1, bd, 43. a ccosa b  cosa, n, n1, n1, ⬁, n!, 44. a n, n1 2, , 755, , 64. Finding the Coefficient of Restitution The coefficient of restitution for steel onto steel is measured by dropping a steel ball, onto a steel plate. If the ball is dropped from a height H and, rebounds to a height h, then the coefficient of restitution is, 1h>H. Suppose that a steel ball is dropped from a height of, 1 m onto a steel plate. Each time the ball hits the plate, it, rebounds to r times it previous height (0  r  1). If the, ball travels a total distance of 2 m, find the coefficient of, restitution for steel on steel., , 34. a 2n5n1, , ⬁, , 2n  1, 35. a, n1 3n  1, , Series, , r 2, r n, r 1, b  Pa1  b  p  Pa1  b  p, 12, 12, 12, , Show that V  12P>r., b. Mrs. Thompson wishes to establish a fund to provide a, university medical center with a monthly research grant, of $150,000. If the fund will earn interest at the rate of, 9% per year compounded monthly, use the result of part, (a) to find the amount of the endowment she is required, to make now., 68. Residual Concentration of a Drug in the Bloodstream Suppose that, a dose of C units of a certain drug is administered to a, patient and that the fraction of the dose remaining in the, patient’s bloodstream t hr after the dose is administered is, given by Cekt, where k is a positive constant., a. Show that the residual concentration of the drug in the, bloodstream after extended treatment when a dose of C, units is administered at intervals of t hr is given by, R, , Cekt, 1  ekt

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756, , Chapter 9 Infinite Sequences and Series, b. If the highest concentration of this particular drug that is, considered safe is S units, find the minimal time that, must exist between doses., Hint: C  R, , S, , 69. Capital Value of a Perpetuity The capital value of a perpetuity, involving payments of P dollars paid at the end of each, investment period into a fund that earns interest at the rate, of r per year compounded continuously is given by, A  Per  Pe2r  Pe3r  p, Find an expression for A that does not involve an infinite, series., 70. Sum of Areas of Nested Squares An infinite sequence of nested, squares is constructed as follows: Starting with a square, with a side of length 2, each square in the sequence is constructed from the preceding square by drawing line segments, connecting the midpoints of the sides of the square. Find the, sum of the areas of all the squares in the sequence., , 72. Prove or disprove: If 兺 an and 兺 bn are both divergent, then, 兺(an  bn) is divergent., 73. Suppose that 兺 an (an, is divergent., , 0) is convergent. Prove that 兺 1>an, , 74. Suppose that 兺 an is convergent and 兺 bn is divergent. Prove, that 兺(an  bn) is divergent., Hint: Prove by contradiction, using Theorem 4., , 75. Suppose that 兺 an is divergent and c, is divergent., , 0. Prove that 兺 can, , Hint: Prove by contradiction, using Theorem 4., , 76. Prove that if the sequence {an} converges, then the series, 兺(an1  an) converges. Conversely, prove that if, 兺(an1  an) converges, then {an} converges., ⬁, 3, 1, 77. Show that a 2 converges and, 2, n1 n, , ⬁, 1, a n2, n1, , 2., , Hint: See Exercise 71 in Section 9.1., ⬁, 1, 78. Prove that a n, converges by showing that {Sn} is, 2, 1, n1, increasing and bounded above, where Sn is the nth partial, sum of the series., , In Exercises 79–84, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, explain why, or give an example to show why it is false., ⬁, 79. If lim n→⬁ an  0, then 兺n1, an converges., , 71. Sum of Areas of Nested Triangles and Circles An infinite sequence, of nested equilateral triangles and circles is constructed as, follows: Beginning with an equilateral triangle with a side of, length 1, inscribe a circle followed by a triangle, followed, by a circle, and so on, ad infinitum. Find the total area of, the shaded regions., , 80. If lim n→⬁ an  L, then the telescoping series, ⬁, 兺n1, (an1  an) converges and has sum L  a1., ⬁, 81. 兺n1, sinn x converges for all x in [0, 2p]., ⬁, ar p, 82. a ar n , provided that 冟 r 冟  1., 1r, np, , 83. If the sequence of partial sums of a series 兺 an is bounded, above, then 兺 an must converge., 84. If 兺(an  bn) converges, then both 兺 an and 兺 bn must, converge., , 9.3, , The Integral Test, The convergence or divergence of a telescoping or geometric series is relatively easy to, determine because we are able to find a simple formula involving a finite number of, terms for the nth partial sum Sn of these series. As we saw in Section 9.2, we can find, the actual sum of a convergent series in this case by simply evaluating lim n→⬁ Sn. However, it is often very difficult or impossible to obtain a simple formula for the nth partial sum of an infinite series, and we are forced to look for alternative ways to investigate the convergence or divergence of the series.

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9.3, , The Integral Test, , 757, , In this and the next two sections we will develop several tests for determining the, convergence or divergence of an infinite series by examining the nth term an of the, series. These tests will confirm the convergence of a series without yielding a value for, its sum. From the practical point of view, however, this is all that is required. Once it, has been ascertained that a series is convergent, we can approximate its sum to any, degree of accuracy desired by adding up the terms of its nth partial sum Sn, provided, that n is chosen large enough. The convergence tests that are given here and in Section 9.4 apply only to series with positive terms., , The Integral Test, ⬁, The Integral Test ties the convergence or divergence of an infinite series 兺n1, an to the, ⬁, convergence or divergence of the improper integral 兰1 f(x) dx, where f(n)  an., , THEOREM 1 The Integral Test, Suppose that f is a continuous, positive, and decreasing function on [1, ⬁). If, f(n)  an for n  1, then, ⬁, , a an, , and, , n1, , 冮, , ⬁, , f(x) dx, , 1, , either both converge or both diverge., , PROOF If you examine Figure 1a, you will see that the height of the first rectangle is, a2  f(2). Since this rectangle has width 1, the area of the rectangle is also a2  f(2)., Similarly, the area of the second rectangle is a3, and so on. Comparing the sum of the, areas of the first (n  1) inscribed rectangles with the area of the region under the, graph of f over the interval [1, n], we see that, n, , 冮 f(x) dx, , a2  a3  p  an, , 1, , which implies that, n, , Sn  a1  a2  a3  p  an, , a1 , , 冮 f(x) dx, , (1), , 1, , y, , y, , y  f (x), , y  f (x), , (1, a1), (2, a2), , (2, a2), (3, a3), (n  1, an1), , (n, an), a3, , a2, 0, , FIGURE 1, , 1, , 2, , (a) a2  a3 , , a4, 3, , a5, 4, , a1, , an, 5, n, , n1 n, ,  an ⱕ y f (x) dx, 1, , x, , 0, , 1, n, , a2, 2, , a3, 3, , a4, 4, , an1, 5, , (b) y f (x) dx ⱕ a1  a2 , 1, , n1 n,  an1, , x

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758, , Chapter 9 Infinite Sequences and Series, , If 兰1⬁ f(x) dx is convergent and has value L, then, n, , 冮 f(x) dx, , a1 , , Sn, , a1  L, , 1, , This shows that {Sn} is bounded above. Also,, Sn1  Sn  an1  Sn, , Because an1  f(n  1)  0, , shows that {Sn} is increasing as well. Therefore, by Theorem 6, Section 9.1, {Sn} is, ⬁, convergent. In other words, 兺n1, an is convergent., Next, by examining Figure 1b, we can see that, n, , 冮 f(x) dx, , a1  a2  p  an1  Sn1, , (2), , 1, , So if 兰1⬁ f(x) dx diverges (to infinity because f(x)  0), then lim n→⬁ Sn1 , ⬁, lim n→⬁ Sn  ⬁ , and 兺n1, an is divergent., Notes, 1. The Integral Test simply tells us whether a series converges or diverges. If it indicates that a series converges, we may not conclude that the (finite) value of the, improper integral used in conjunction with the test is the sum of the convergent, series (see Exercise 54)., 2. Since the convergence of an infinite series is not affected by adding or subtracting, ⬁, a finite number of terms to the series, we sometimes study the series 兺nN, an , ⬁, p, aN  aN1  rather than the series 兺n1 an. In this case the series is compared, with the improper integral 兰N⬁ f(x) dx, as we will see in Example 2., ⬁, , 1, , EXAMPLE 1 Use the Integral Test to determine whether a 2, converges or, n1 n  1, diverges., Solution Here, an  f(n)  1>(n 2  1), so we consider the function f(x) , 1>(x 2  1). Since f is continuous, positive, and decreasing on [1, ⬁), we may use the, Integral Test. Next,, , 冮, , 1, , ⬁, , 1, x 1, 2, , dx  lim, , 冮, , b→⬁ 1, , b, , 1, x 1, 2, , dx  lim Ctan1 xD 1, b, , b→⬁, ,  lim (tan1 b  tan1 1) , b→⬁, , p, p, p,  , 2, 4, 4, , ⬁, Since 兰 1>(x  1) dx converges, we conclude that 兺n1, 1>(n 2  1) converges as, well., ⬁, 1, , 2, , ⬁, , ln n, , EXAMPLE 2 Use the Integral Test to determine whether a, converges or, n1 n, diverges., Solution Here, an  (ln n)>n, so we consider the function f(x)  (ln x)>x. Observe, that f is continuous and positive on [1, ⬁). Next, we compute, , f ¿(x) , , 1, xa b  ln x, x, x, , 2, , , , 1  ln x, x2

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9.3, , The Integral Test, , 759, , Note that f ¿(x)  0 if ln x  1, that is, if x  e. This shows that f is decreasing on, [3, ⬁). Therefore, we may use the Integral Test:, , 冮, , 3, , ⬁, , ln x, dx  lim, x, b→⬁,  lim, , b→⬁, , 冮, , b, , 3, , b, ln x, 1, dx  lim c (ln x)2 d, x, b→⬁ 2, 3, , 1, [(ln b) 2  (ln 3)2]  ⬁, 2, , ⬁, and we conclude that 兺n1, (ln n)>n diverges., , The p-Series, The following series will play an important role in our work later on., , DEFINITION p-Series, A p-series is a series of the form, ⬁, , 1, 1, 1, p, a n p  1  2p  3p , n1, where p is a constant., ⬁, Observe that if p  1, the p-series is just the harmonic series 兺n1, 1>n., The conditions for the convergence or divergence of the p-series can be found by, applying the Integral Test to the series., , THEOREM 2 Convergence of the p-Series, ⬁, 1, The p-series a p converges if p  1 and diverges if p, n, n1, , 1., , PROOF If p  0, then lim n→⬁ (1>n p)  ⬁ . If p  0, then lim n→⬁ (1>n p)  1. In either, , case, lim n→⬁ (1>n p) 0, so the p-series diverges by the Divergence Test., If p  0, then the function f(x)  1>x p is continuous, positive, and decreasing on, [1, ⬁). In Example 2 in Section 7.6 we found that 兰1⬁ 1>x p dx converges if p  1 and, ⬁, diverges if p 1. Using this result and the Integral Test, we conclude that 兺n1, 1>n p, ⬁, p, converges if p  1 and diverges if 0  p 1. Therefore, 兺n1 1>n converges if p  1, and diverges if p 1., , EXAMPLE 3 Determine whether the given series converges or diverges., ⬁, , a. a, n1, , 1, n2, , ⬁, 1, b. a, n1 1n, , ⬁, , c. a n 1.001, n1, , Solution, a. This is a p-series with p  2  1, and hence it converges by Theorem 2., ⬁, b. Rewriting the series in the form 兺n1, 1>n 1>2, we see that the series is a p-series, 1, with p  2  1, and hence it diverges by Theorem 2., ⬁, c. We rewrite the series in the form 兺n1, 1>n 1.001, which we recognize to be a, p-series with p  1.001  1 and conclude that the series converges.

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760, , Chapter 9 Infinite Sequences and Series, , 9.3, , CONCEPT QUESTIONS, , ⬁, 1, 1, 1, 1, 1, 1, 1. Consider the series a 2  2  2  2  2  2  p ., 1, 2, 3, 4, 5, n1 n, 1, Let f(x)  2 ., x, a. Sketch a figure similar to Figure 1a for this series and, function, and compute a1  f(1) , a2  f(2), a3  f(3),, p , an  f(n)., 1, 1, 1, 1, b. Explain why Sn  2  2  2  p  2, 1, 2, 3, n, n, ⬁, 1, 1, 1, , dx  1 , dx., 2, 2, 12, 1 x, 1 x, c. By evaluating the improper integral in part (b), show that, Sn 2 for each n  1, 2, 3, p . Then use the Monotone, ⬁, 1, Convergence Theorem (Section 9.1) to show that a 2, n1 n, converges., , 冮, , ⬁, 1, 1, 1, 1, 1, 1, 2. Consider the series a       p ., 1, 2, 3, 4, 5, n1 n, 1, Let f(x)  ., x, a. Sketch a figure similar to Figure 1b for this series and, function, and compute a1  f(1) , a2  f(2) , a3  f(3) ,, p , an  f(n) ., n, 1, 1, 1, 1, 1, b. Explain why Sn1     p , dx., , 1, 2, 3, n1, 1 x, ⬁, ⬁, 1, 1, c. Show that, dx is divergent, and conclude that a, x, n1 n, 1, diverges., , 冮, , 冮, , 冮, , Note: This is the harmonic series that was shown to be divergent in, Section 9.2., , Note: The Swiss mathematician Leonhard Euler showed that the sum, of this series is p2>6., , 9.3, , EXERCISES, , In Exercises 1–8, use the Integral Test to determine whether the, series is convergent or divergent., ⬁, , ⬁, , 1, , 3, 2. a, n1 2n  1, , 1. a 4, n1 n, ⬁, , 21. a, , n1, , n2, ⬁, , 1, 1, 1, 1, 1, , , p, 5.  , 2, 5, 10, 17, 26, , n1, , (n  1), , ⬁, ⬁, , n, 2, , 3>2, , 1, 8. a, n2 n 1ln n, , In Exercises 9–14, determine whether the p-series is convergent, or divergent., ⬁, 1, 9. a 3, n1 n, , ⬁, , 10. a, , 1, , ⬁, 1, 11. a 1.01, n1 n, , n 2>3, ⬁, 1, 12. a e, n, n1, , ⬁, , ⬁, , 13. a n p, n1, , n1, , 14. a n 0.98, n1, , In Exercises 15–32 determine whether the given series is, convergent or divergent., ⬁, , 1, 15. a, n0 1n  1, , ln n, n, 1, , 23. a, 2, n2 n(ln n), , 1, 1, 1, 1, 1,  , , , p, 3, 7, 11, 15, 19, , 7. a, , 1, 2,  2b, 19. a a, n, n1 n1n, ⬁, , 4. a nen, , n1, , ⬁, , ⬁, , ⬁, , 3. a en, , 6., , ⬁, 1, 17. a, n1 n1n, , ⬁, , 16. a, n1, , n, 22n 2  1, , 25. a, n1, ⬁, , 1, sina b, n, n2, 1, , ⬁, , 18. a n 0.75, n1, ⬁, 2 n, 1, 20. a c a b  3>2 d, 3, n1, n, ⬁, ln n, 22. a 2, n, n2, , ⬁, e1>n, 24. a 2, n1 n, , ⬁, 1, 26. a, 1n, 4, n1, , 27. a, 2, n1 4n  1, , ⬁, n, 28. a n, 2, n1, , ⬁, tan1 n, 29. a 2, n1 n  1, , ⬁, 1, 30. a n, e, 1, n1, , ⬁, 1, 31. a 2, n1 n  2n  5, , ⬁, 1, 32. a, 2, n1 2n  7n  3, , In Exercises 33 and 34, find the values of p for which the series, is convergent., ⬁, 1, 33. a, p, n2 n(ln n), , ⬁, ln n, 34. a p, n1 n, , 35. Find the value(s) of a for which the series, ⬁, 1, a, a c n  1  n  2 d converges. Justify your answer., n1, , V Videos for selected exercises are available online at www.academic.cengage.com/login.

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9.3, 36. a. Show that if Sn is the nth partial sum of the harmonic, series, then Sn 1  ln n., Hint: Use Inequality (1), page 757, with f(x)  1>x., , b. Use part (a) to show that the sum of the first 1,000,000, terms of the harmonic series is less than 15. The harmonic series diverges very slowly!, , 1, , Sn , , 3, 2, , and therefore,, 0  ln(n  1)  ln n, , Hence, deduce that the sequence {an} defined by, an  1 , , 1 p 1,    ln n, n, 2, , is bounded below., Hint: Use Inequality (2), page 758, with f(x)  1>x., , b. Show that, , 冮, , 1, , n1, , n, , n1, , 1, dx  ln(n  1)  ln n, x, , Hint: Draw a figure similar to Figure 1., , ⬁, 1, 43. a 2, ;, n, 1, n1, , 1, 1, g  lim an  lim a1   p   ln nb, n→⬁, n→⬁, 2, n, whose value is 0.5772 p , is called Euler’s constant., , 38. Riemann Zeta Function The Riemann zeta function for real, numbers is defined by, ⬁, , j(x)  a n x, , 1, n, , ⬁, , 1, a n2, n1, , 2, , In Exercises 41–44, use the result of Exercise 39 to find the maximum error if the sum of the series is approximated by Sn., ⬁, 2, 41. a 2 ;, n1 n, , Note: The number, , Sn , , Confirm this result, using the result of part (a)., c. Use the result of Exercise 39a to find the upper and, lower bounds on the error incurred in approximating, ⬁, 1, a n 2 using the 100th partial sum of the series., n1, ⬁, 1, p2, cas d. It can be shown that a, . Use a calculator or, , 2, 6, n1 n, computer to verify this., , and use this result to show that the sequence {an}, defined in part (a) is decreasing., c. Use the Monotone Convergence Theorem to show that, {an} is convergent., , ⬁, 1, a n2, n1, , 1, n1, , n, ⬁, 1, 1, where Sn  a 2 is the nth partial sum of a 2 ., k1 k, n1 n, b. In Exercise 77 in Section 9.2 you were asked to show, that, , 1 p 1,  , n, 2, , 1, 1, 1   p   ln n, n, 2, , 761, , ⬁, 1, 40. Consider the series a 2 , which is a convergent p-series, n1 n, (p  2) ., a. Use the result of Exercise 39b to show that, , 37. Euler’s Constant, a. Show that, ln(n  1), , The Integral Test, , ⬁, , 42. a, , S40, , 1, , ;, , S20, , 44. a nen ;, , S3, , n1, , n 5>2, , ⬁, , S50, , 2, , n1, , In Exercises 45–48, use the result of Exercise 39 to find the number of terms of the series that is sufficient to obtain an approximation of the sum of the series accurate to two decimal places., ⬁, 1, 45. a 2, n1 n, , ⬁, 1, 46. a 3, n1 n, , ⬁, tan1 n, 47. a, 2, n1 1  n, , ⬁, 1, 48. a, 2, n2 n(ln n), , n1, , What is the domain of the function?, 39. Let ak  f(k), where f is a continuous, positive, and, ⬁, an, decreasing function on [n, ⬁) , and suppose that 兺n1, is convergent., a. Show, by sketching appropriate figures, that if, ⬁, Rn  S  Sn, where S  兺n1, an and Sn  兺nk1 ak, then, , 冮, , ⬁, , f(x) dx, , Rn, , n1, , 冮, , f(x) dx, , n, , Note: Rn is the error estimate for the Integral Test., , 冮, , ⬁, , n1, , f(x) dx, , S, , ⬁, 1, 49. a 4, n1 n, , ⬁, , 50. a, n1, , 1, n 9>2, , 51. a. Show that, ⬁, , ⬁, , b. Use the result of part (a) to deduce that, Sn , , In Exercises 49 and 50, use the result of Exercise 39 to find the, sum of the series accurate to three decimal places using the nth, partial sum of the series., , Sn , , 冮, , n, , ⬁, , f(x) dx, , ⬁, 1, 1, 1, , a n(n  1)(n  2), a c 2n(n  1)  2(n  1)(n  2) d, n1, n1, , b. Use the results of part (a) to evaluate, ⬁, 1, a n(n  1)(n  2), n1

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762, , Chapter 9 Infinite Sequences and Series, , ⬁, 1, 52. Evaluate a 3 accurate to four decimal places by estabn1 n, lishing parts (a) and (b) and using the results of Exercise 51., , b. Show that the given series is a geometric series, and find, its sum., c. Conclude that although the convergence of 兰0⬁ ex dx, implies convergence of the infinite series, its value does, not give the sum of the infinite series., , ⬁, ⬁, ⬁, 1, 1, 1, a. a 3  1  a,  a 3 2, (n, , 1)n(n, , 1), n, n, (n,  1), n1, n2, n2, ⬁, 1, b. a 3 2, can be approximated with an accuracy of, n2 n (n  1), four decimal places by using six terms of the series., , Hint: Show that, 1, , 2, , n 3 (n 2  1), , n5, , if n  2, and use the result of Exercise 39., ⬁, 1, 53. Use the Integral Test to show that a, n(ln, n)[ln(ln, n)]p, n3, converges if p  1 and diverges if p 1., ⬁, 兺n0, , n, , e ., 54. Consider the series, a. Evaluate 兰0⬁ ex dx, and deduce from the Integral Test, that the given series is convergent., , 9.4, , In Exercises 55–58, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, explain why, or give an example to show why it is false., 55. Suppose that f is a continuous, positive, and decreasing, ⬁, function on [1, ⬁). If f(n)  an for n  1 and 兺n1, an is, ⬁, ⬁, convergent, then 兺n1 an a1  兰1 f(x) dx., 56. Suppose that f is a continuous, positive, and decreasing, function on [1, ⬁). If f(n)  an for n  1 and, ⬁, 兰N⬁ f(x) dx  ⬁ , where N is a positive integer, then 兺n1 an, diverges., 57., , 冮, , 1, , ⬁, , dx, ⬁, x(x  1), , ⬁, 58. If 兺n1, an is a convergent series with positive terms, then, ⬁, 兺n1 1an must also converge., , The Comparison Tests, The rationale for the comparison tests is that the convergence or divergence of a given, series 兺 an can be determined by comparing its terms with the corresponding terms of, a test series whose convergence or divergence is known. The series that we will consider in this section have positive terms., , an, b1, , The Comparison Test, , b3, , b2, , b4, b5, , b6, , a1 a2 a3 a4 a5 a6, 0, , 1, , 2, , 3, , 4, , 5, , 6, , 7, , n, , FIGURE 1, Each rectangle representing an is contained in the rectangle representing bn., , Suppose that the terms of a series 兺 an are smaller than the corresponding terms of a, series 兺 bn. This situation is illustrated in Figure 1, where the respective terms of each, series are represented by rectangles, each of width 1 and having the appropriate height., If 兺 bn is convergent, the total area of the rectangles representing this series is finite., Since each rectangle representing the series 兺 an is contained in a corresponding rectangle representing the terms of 兺 bn, the total area of the rectangles representing 兺 an, must also be finite; that is, the series 兺 an must be convergent. A similar argument suggests that if all the terms of a series 兺 an are larger than the corresponding terms of a, series 兺 bn that is known to be divergent, then 兺 an must itself be divergent. These, observations lead to the following theorem., , THEOREM 1 The Comparison Test, Suppose that 兺 an and 兺 bn are series with positive terms., a. If 兺 bn is convergent and an bn for all n, then 兺 an is also convergent., b. If 兺 bn is divergent and an  bn for all n, then 兺 an is also divergent.

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9.4, , The Comparison Tests, , 763, , PROOF Let, n, , n, , Sn  a ak, , Tn  a bk, , and, , k1, , k1, , be the nth terms of the sequence of partial sums of 兺 an and 兺 bn, respectively. Since, both series have positive terms, {Sn} and {Tn} are increasing., ⬁, a. If 兺n1, bn is convergent, then there exists a number L such that lim n→⬁ Tn  L and, Tn L for all n. Since an bn for all n, we have Sn Tn, and this implies that, Sn L for all n. We have shown that {Sn} is increasing and bounded above, so by, the Monotone Convergence Theorem for Sequences of Section 9.1, 兺 an converges., ⬁, b. If 兺n1, bn is divergent, then lim n→⬁ Tn  ⬁ , since {Tn} is increasing. But an  bn, for all n, and this implies that Sn  Tn, which in turn implies that lim n→⬁ Sn  ⬁ ., Therefore, 兺 an diverges., , To use the Comparison Test, we need a catalog of test series whose convergence, or divergence is known. For the moment we can use the geometric series and the pseries as test series., ⬁, , 1, , EXAMPLE 1 Determine whether the series a 2, converges or diverges., n1 n  2, Solution, , Let, an , , 1, n 2, 2, , If n is large, n  2 behaves like n , so an behaves like, 2, , 2, , bn , , 1, n2, , This observation suggests that we compare 兺 an with the test series 兺 bn, which is a, convergent p-series with p  2. Now,, 0, , 1, n 2, 2, , , , 1, , n1, , n2, , and the given series is indeed “smaller” than the test series 兺 1>n 2. Since the test series, converges, we conclude by the Comparison Test that 兺 1>(n 2  2) also converges., ⬁, , 1, , EXAMPLE 2 Determine whether the series a, n converges or diverges., n1 3  2, Solution, , Let, an , , 1, 3  2n, , If n is large, 3  2n behaves like 2n, so an behaves like bn  1 12 2 n. This observation sugn, gests that we compare 兺 an with 兺 bn. Now the series 兺 21n  兺 1 12 2 is a geometric series, 1, with r  2  1, so it is convergent. Since, an , , 1, 1,  n  bn, 3  2n, 2, , n1, , the Comparison Test tells us that the given series is convergent.

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764, , Chapter 9 Infinite Sequences and Series, , Note Since the convergence or divergence of a series is not affected by the omission, of a finite number of terms of the series, the condition an bn (or an  bn) for all n, can be replaced by the condition that these inequalities hold for all n  N for some, integer N., ⬁, , 1, , EXAMPLE 3 Determine whether the series a, is convergent or divergent., n2 1n  1, Solution, , Let, an , , 1, 1n  1, , If n is large, 1n  1 behaves like 1n, so an behaves like, bn , , 1, 1n, , Now the series, ⬁, , ⬁, ⬁, 1, 1, b, , , a n, a 1n, a 1>2, n2, n2, n2 n, , is a p-series with p  12  1, so it is divergent. Since, an , , 1, 1, ,  bn, 1n  1, 1n, , for n  2, , the Comparison Test implies that the given series is divergent., , The Limit Comparison Test, Consider the series, ⬁, 1, a 1n  1, n1, , If n is large, 1n  1 behaves like 1n, so the nth term of the given series, an , , 1, 1n  1, , behaves like, bn , , 1, 1n, , ⬁, ⬁, bn  兺n1, 1> 1n is a divergent p-series with p  12, we expect, Since the series 兺n1, ⬁, the series 兺n1 1>( 1n  1) to be divergent as well. But the inequality, , an , , 1, 1, ,  bn, 1n  1, 1n, , n1, , ⬁, an is “smaller” than a divergent series, and this is of no help if we, tells us that 兺n1, try to use the Comparison Test!, In situations like this, the Limit Comparison Test might be applicable. The rationale for this test follows: Suppose that 兺 an and 兺 bn are series with positive terms and

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9.4, , The Comparison Tests, , 765, , suppose that lim n→⬁ (an>bn)  L, where L is a positive constant. If n is large, an>bn ⬇ L, or an ⬇ Lbn. It is reasonable to conjecture that the series 兺 an and 兺 bn must both converge or both diverge., , THEOREM 2 The Limit Comparison Test, Suppose that 兺 an and 兺 bn are series with positive terms and, lim, , n→⬁, , an, L, bn, , where L is a positive number. Then either both series converge or both diverge., , PROOF Since lim n→⬁ (an>bn)  L  0, there exists an integer N such that n  N, implies that, , `, , an, 1,  L`  L, bn, 2, , an, 1, 3, L,  L, 2, bn, 2, or, 1, 3, Lbn  an  Lbn, 2, 2, If 兺 bn converges, so does 兺 32 Lbn. Therefore, the right side of the last inequality implies, that 兺 an converges by the Comparison Test. On the other hand, if 兺 bn diverges, so, does 兺 12 Lbn, and the left side of the last inequality implies by the Comparison Test, that 兺 an diverges as well., ⬁, , 1, , EXAMPLE 4 Show that the series a, is divergent., n1 1n  1, Solution As we saw earlier, 1>( 1n  1) behaves like 1> 1n if n is large. This suggests that we use the Limit Comparison Test with an  1>( 1n  1) and bn  1> 1n., Thus,, 1, an, 1n, 1, 1n  1, lim,  lim,  lim,  lim, 1, n→⬁ bn, n→⬁, 1, n→⬁ 1n  1, n→⬁, 1, 1, 1n, 1n, , ⬁, Since 兺n1, 1> 1n is divergent 1 it is a p-series with p  12 2 , we conclude that the given, series is divergent as well., , Note, , You can still use the Comparison Test to solve the problem. Simply observe that, 1, 1, 1, , , 1n  1, 1n  1n, 21n, , for n  1, , This suggests picking 兺 bn, where bn  1>(2 1n), for the test series.

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766, , Chapter 9 Infinite Sequences and Series, ⬁, , 2n 2  n, , EXAMPLE 5 Determine whether the series a, n1, , Solution, fore,, , 24n 7  3, , converges or diverges., , If n is large, 2n 2  n behaves like 2n 2, and 4n 7  3 behaves like 4n 7. There2n 2  n, , an , , 24n 7  3, , behaves like, 2n 2, 24n, , 7, , , , 2n 2, 2n, , , , 7>2, , 1, n 3>2, ,  bn, , Now, lim, , n→⬁, , an, 2n 2  n, n 3>2,  lim, ⴢ, bn n→⬁ (4n 7  3)1>2, 1,  lim, , n→⬁, , 2n 7>2  n 5>2, (4n 7  3)1>2, 1, n, , 2,  lim, , n→⬁, , a4 , , b, 7, , 3, n, , 1>2, , Divide numerator and, denominator by n 7>2., , 1, , Since 兺 1>n 3>2 converges 1 it is a p-series with p  32 2 , the given series converges, by, the Limit Comparison Test., ⬁, , EXAMPLE 6 Determine whether the series a, , 1n  ln n, n2  1, , n1, , converges or diverges., , Solution If n is large, 1n  ln n behaves like 1n. You can see this by comparing, the derivatives of f(x)  1x and t(x)  ln x:, f ¿(x) , , 1, 21x, , t¿(x) , , and, , 1, x, , Observe that t¿(x) approaches zero faster than f ¿(x) approaches zero, as x → ⬁ . This, shows that 1x grows faster than ln x. Also, if n is large, n 2  1 behaves like n 2. Therefore,, an , , 1n  ln n, n2  1, , behaves like, 1n, n, , 2, , , , 1, n 3>2, ,  bn

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9.4, , The Comparison Tests, , 767, , Next, we compute, lim, , n→⬁, , an, n 1>2  ln n n 3>2,  lim, ⴢ, bn n→⬁ n 2  1, 1,  lim, , n 2  n 3>2 ln n, n2  1, , n→⬁, , 1,  lim, , n→⬁, , 1, , ln n, n 1>2, 1, , Divide the numerator and, denominator by n 2., , n2, , In evaluating this limit, we need to compute, , lim, , x→⬁, , ln x, x 1>2, , 1, x, ,  lim, , x→⬁ 1 x 1>2, ,  lim, , 2, , x→⬁, , 2, 0, 1x, , Use l’Hôpital’s Rule., , (Incidentally, this result supports the observation made earlier that 1x grows faster, than ln x.) Using this result, we find, an, lim,  lim, n→⬁ bn, n→⬁, , 1, 1, , ln n, n 1>2, 1, 1, n2, , Since 兺 1>n 3>2 converges 1 it is a p-series with p  32 2 , the given series converges, by, the Limit Comparison Test., , 9.4, , CONCEPT QUESTIONS, , 1. a. State the Comparison Test and the Limit Comparison, Test., b. When is the Comparison Test used? When is the Limit, Comparison Test used?, 2. Let 兺 an and 兺 bn be series with positive terms., a. If 兺 bn is convergent and an  bn for all n, what can you, say about the convergence or divergence of 兺 an? Give, examples., b. If 兺 bn is divergent and an bn for all n, what can you, say about the convergence or divergence of 兺 an? Give, examples., , 9.4, , In Exercises 3 and 4, let 兺 an , 兺 bn , and 兺 cn be series with positive terms., 3. If 兺 an is convergent and bn  cn an for all n, what can, you say about the convergence or divergence of 兺 bn and, 兺 cn?, 4. If 兺 an is divergent and bn  cn  an for all n, what can you, say about the convergence or divergence of 兺 bn and 兺 cn?, , EXERCISES, , In Exercises 1–12, use the Comparison Test to determine, whether the series is convergent or divergent., ⬁, 1, 1. a, 2, 2n, 1, n1, , ⬁, 1, 2. a 2, n, , 2n, n1, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , ⬁, 1, 3. a, n3 n  2, ⬁, , 5. a, n2, , 1, 2n  1, 2, , ⬁, , 4. a, n2, ⬁, , 6. a, n0, , 1, n 2>3  1, 1, 2n  1, 3

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768, , Chapter 9 Infinite Sequences and Series, , ⬁, 2n, 7. a n, n0 3  1, ⬁, , 9. a, n2, , ⬁, 3n, 8. a n, n3 2  4, ⬁, , ln n, n, , 10. a, n1, , ⬁, , 2  sin n, 11. a, 3n, n1, , cos2 n, n2, , ⬁, , 1, 12. a n, n1 n, , In Exercises 13–24, use the Limit Comparison Test to determine, whether the series is convergent or divergent., ⬁, n, 13. a 2, n2 n  1, ⬁, , 15. a, n2, , n, 2n 5  1, , ⬁, , 3n 2  1, 17. a, 5, n1 2n  n  2, ⬁, , 19. a, n2, , 1, 2n  n  1, 3, , ⬁, n, 21. a n, 2, 1, n1, ⬁, , 1, 23. a sin, n, n1, , ⬁, 1, 14. a, n1 1n  2, ⬁, 2n  1, 16. a, 2, n1 3n  n  1, ⬁, n2  1, 18. a 2, n1 n (n  3), ⬁, 1, 20. a n, n2 2  3, ⬁, ln n, 22. a 3, n, 1, n2, ⬁, , 1, 24. a tan, n, n1, , In Exercises 25–40, determine whether the series is convergent, or divergent., ⬁, n1, 25. a, 2, n1 (n  2)(2n  1), , 26. a, , ⬁, n1, 27. a 3, n1 n  2, , ⬁, n1, 28. a, 3, n1 2n  1, , ⬁, 2n1, 29. a 2, n1 n  n, , 30. a, , ⬁, sin2 n, 31. a, n1 n 1n  1, , ⬁, tan1n, 32. a 3, n1 n  1, , ⬁, , 1, 33. a, n2 ln n, ⬁, , 1, 35. a, n0 n!, ⬁, , n!, 37. a n, n, n1, ⬁, , 39. a, n1, , 1n  ln n, 2n 2  3, , ⬁, n1, , n, , n1, , 1, n  2n 2  1, , ⬁, , ln n, 34. a, n1 n  2, ⬁, , 1, 38. a, 1, , 2, , 3, pn, n1, n1, , ⬁, 1, 44. a n, 3, 1, n1, , 45. a, , ⬁, n1, ⬁, n1, , sin n  2, n4, tan1 n, 2n, , 46. Suppose that 兺 an is a convergent series with positive terms., Show that 兺(an>n) is also convergent., 47. Suppose that 兺 an and 兺 bn are convergent series with positive terms. Show that 兺 anbn is convergent., Hint: There exists an integer N such that n  N implies that bn, and therefore, anbn, , 1,, , an for n  N., , 48. Suppose that 兺 an is a convergent series with positive terms, and {cn} is a sequence of positive numbers that converges to, zero. Prove that 兺 ancn is convergent., Hint: There exists an integer N such that n  N implies that cn  L,, where L is a positive number, and therefore, ancn  Lan for n  N., , 49. Prove that if an  0 and 兺 an converges, then 兺 a 2n also converges. Is the converse true? Explain., 50. Using the result of Exercise 48 or otherwise, show that, ⬁, 兺n2, 1>(n p ln n) is convergent if p  1., 51. a. Suppose that 兺 an and 兺 bn are series with positive terms, and 兺 bn is convergent. Show that if lim n→⬁ an>bn  0,, then 兺 an is convergent., ⬁, ln n, b. Use part (a) to show that a 2 is convergent., n, 52. Give an example of a pair of series 兺 an and 兺 bn with, positive terms such that lim n→⬁ an>bn  0, 兺 bn is divergent,, but 兺 an is convergent. (Compare this with the result of, Exercise 51.), 53. a. Show that if 兺 an is a convergent series with positive, terms, then 兺 sin an is also convergent., b. If 兺 an diverges can 兺 sin an converge? Explain., 54. Prove that (a), (b), , 冮, , 1, , ⬁, , ⬁, , 43. a, , 2, , n, 36. a, n1 n!, , 40. a, , ⬁, 1, 42. a 3, n, , 2n, n1, , n1, , 2n  n, 5, , ⬁, , In Exercises 42–45, use the result of Exercise 41 to find an, approximation of the sum of the series using its partial sum,, accurate to two decimal places., , ⬁, , 冮, , 1, , ⬁, , 1, dx converges and, 1x(x  1)(x  2), , 1, dx diverges., 1x(x  1), , In Exercises 55–58, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, explain why, or give an example to show why it is false., , 2n 2  n, , 55. If 0, , an, , bn and 兺 an converges, then 兺 bn diverges., , 23n 7  ln n, , 56. If 0, , an, , bn and 兺 bn diverges, then 兺 an diverges., , 41. Let 兺 an and 兺 bn be series with 0 an bn, and suppose, that 兺 bn is convergent with sum T. Then the Comparison, Test implies that 兺 an also converges, say, with sum S. Put, Rn  S  Sn and Tn  T  Un, where Un is the nth-partial, sum of 兺 bn. Show that the remainders Rn and Tn satisfy, Rn Tn., , 57. If an  0 and bn  0 and 兺 anbn converges, then 兺 an and, 兺 bn both converge., 58. If an  0 and bn  0 and 兺 2a 2n  b 2n converges, then 兺 an, and 兺 bn both converge.

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9.5, , 9.5, , Alternating Series, , 769, , Alternating Series, Up to now, we have dealt mainly with series that have positive terms, and the convergence tests that we have developed are applicable only to these series. In this section, and Section 9.6 we will consider series that contain both positive and negative terms., Series whose terms alternate in sign are called alternating series., Examples are the alternating harmonic series, ⬁, , (1)n1, 1, 1, 1, 1, 1, 1     p, a, n, 2, 3, 4, 5, 6, n1, and the series, ⬁, , (1)n n 2, 1, 4, 9, 16, 25 p, a (n  1)!   2!  3!  4!  5!  6! , n1, More generally, an alternating series is a series of the form, ⬁, , ⬁, , n1, a (1) an, , or, , n1, , n, a (1) an, n1, , where an is a positive number. We use the Alternating Series Test to determine convergence for these series., , THEOREM 1 The Alternating Series Test, If the alternating series, ⬁, , a (1), , an  a1  a2  a3  a4  a5  a6  p, , n1, , an  0, , n1, , satisfies the conditions, 1. an1 an for all n, 2. lim an  0, n→⬁, , then the series converges., , The plausibility of Theorem 1 is suggested by Figure 1, which shows the first few, terms of the sequence of partial sums {Sn} of the alternating series, , a1, a2, , ⬁, , a3, , a (1), , a4, a5, 0, , S2 S4, , S, , S5, , an  a1  a2  a3  p, , n1, , n1, , S3, , S1, , FIGURE 1, The terms of {Sn} oscillate in smaller, and smaller steps, and this suggests that, lim n→⬁ Sn  S., , plotted on the number line. The point S2  a1  a2 lies to the left of the number, S1  a1, since it is obtained by subtracting the positive number a2 from S1. But the, number S2 also lies to the right of the origin because a2 a1. The number, S3  a1  a2  a3  S2  a3 is obtained by adding a3 to S2, and hence it lies to the, right of S2. But because a3  a2, S3 lies to the left of S1. Continuing in this fashion,, we see that numbers corresponding to the partial sums {Sn} oscillate. Because, lim n→⬁ an  0, the steps get smaller and smaller. Thus, it appears that the sequence, {Sn} will approach a limit. In particular, observe that the even terms of the sequence, {Sn} are increasing, whereas the odd terms of the sequence are decreasing. This suggests that the subsequence {S2n} will approach the limit S from below and the subsequence {S2n1} will approach S from above. These observations form the basis of the, proof of Theorem 1.

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9.5, , Alternating Series, , 771, , EXAMPLE 2 Determine whether the series converges or diverges., ⬁, 2n, a. a (1)n, 4n  1, n1, , ⬁, , b. a (1)n1, n1, , 3n, 4n  1, 2, , Solution Since both series are alternating series we use the Alternating Series Test., a. Here, an  2n>(4n  1). Because, 2n, 1, , 4n  1, 2, , lim, , n→⬁, , 0, , we see that condition (2) in the Alternating Series Test is not satisfied. In fact,, this computation shows that, lim (1)n, , n→⬁, , 2n, 4n  1, , does not exist, and the divergence of the series follows from the Divergence, Test., b. Here an  3n>(4n 2  1). First we show that an  an1 for all n. We can do this, by showing that f(x)  3x>(4x 2  1) is decreasing for x  0. We compute, f ¿(x) , , , (4x 2  1)(3)  (3x)(8x), (4x 2  1)2, 12x 2  3, (4x 2  1)2, , 0, , and the desired conclusion follows. Next, we compute, , lim an  lim, , n→⬁, , n→⬁, , 3n, 4n 2  1, ,  lim, , n→⬁, , 3, n, 4, , 1, , 0, , n2, , Since both conditions of the Alternating Series Test are satisfied, we conclude, that the series is convergent., Notes, 1. Example 2a reminds us once again that it is a good idea to begin investigating, the convergence of a series by checking for divergence using the Divergence, Test., 2. Because the behavior of a finite number of terms will not affect the convergence, or divergence of a series, the first condition in the Alternating Series Test can be, replaced by the condition an1 an for n  N, where N is some positive integer., , Approximating the Sum of an Alternating Series by Sn, Suppose that we can show that the series 兺 an is convergent so that it has a sum S. If, {Sn} is the sequence of partial sums of 兺 an, then lim n→⬁ Sn  S or, equivalently,, lim (S  Sn)  0, , n→⬁

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772, , Chapter 9 Infinite Sequences and Series, , Thus, the sum of a convergent series can be approximated to any degree of accuracy, by its nth partial sum Sn, provided that n is taken large enough. To measure the accuracy of the approximation, we introduce the quantity, ⬁, , n, , k1, , k1, , ⬁, , Rn  S  Sn  a ak  a ak  a ak  an1  an2  an3  p, kn1, , ⬁, called the remainder after n terms of the series 兺n1, an. The remainder measures the, error incurred when S is approximated by Sn., In general, it is difficult to determine the accuracy of such an approximation, but, for alternating series the following theorem gives us a simple way of estimating the, error., , THEOREM 2 Error Estimate in Approximating an Alternating Series, ⬁, Suppose 兺n1, (1)n1 an is an alternating series satisfying, , 1. 0 an1 an for all n, 2. lim an  0, n→⬁, , If S is the sum of the series, then, 冟 Rn 冟  冟 S  Sn 冟, , an1, , In other words, the absolute value of the error incurred in approximating S by, Sn is no larger than an1, the first term omitted., , PROOF We have, ⬁, , n, , k1, , k1, , ⬁, , S  Sn  a (1)k1ak  a (1)k1ak  a (1)k1ak, kn1, ,  (1)nan1  (1)n1an2  (1)n2an3  p,  (1)n(an1  an2  an3  p), Next,, an1  an2  an3  an4  p,  (an1  an2)  (an3  an4)  p, 0, , Since an1, , So, 冟 S  Sn 冟  an1  an2  an3  an4  an5  p,  an1  (an2  an3)  (an4  an5)  p, Since every expression within each parenthesis is nonnegative, we see that, 冟 S  Sn 冟 an1., , !, , This error estimate holds only for alternating series., , an for all n

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9.5, ⬁, , Alternating Series, , 773, , 1, , EXAMPLE 3 Show that the series a (1)n is convergent, and find its sum corn!, n0, , rect to three decimal places., Solution, , Since, an1 , , 1, 1, 1, , ,  an, (n  1)!, n!(n  1), n!, , for all n and, lim an  lim, , n→⬁, , n→⬁, , 1, 0, n!, , we conclude that the series converges by the Alternating Series Test., To see how many terms of the series are needed to ensure the specified accuracy, of the approximation, we turn to Theorem 2. It tells us that, 冟 Rn 冟  冟 S  Sn 冟, , an1 , , 1, (n  1)!, , We require that 冟 Rn 冟  0.0005, which is satisfied if, 1,  0.0005, (n  1)!, , or, , (n  1)! , , 1,  2000, 0.0005, , The smallest positive integer that satisfies the last inequality is n  6. Hence, the, required approximation is, S ⬇ S6 , , 1, 1, 1, 1, 1, 1, 1, , , , , , , 0!, 1!, 2!, 3!, 4!, 5!, 6!, , 11, , 1, 1, 1, 1, 1,  , , , 2, 6, 24, 120, 720, , ⬇ 0.368, , 9.5, , CONCEPT QUESTIONS, , 1. a. What is an alternating series? Give an example., b. State the Alternating Series Test, and use it to determine, whether the series in your example converges or, diverges., , 9.5, , EXERCISES, , In Exercises 1–24, determine whether the series converges or, diverges., ⬁, , (1)n1, 1. a, n1 n  2, ⬁, , 3. a, n1, , c. What is the maximum error that can occur if you approximate the sum of a convergent alternating series by its, nth partial sum?, , (1)n1, n2, , ⬁, , (1)nn, 2. a, n1 3n  1, ⬁, , 4. a, n1, , (1)n1n 2, 2n 2  1, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , ⬁, , 5. a, n1, ⬁, , 7. a, n2, ⬁, , (1)n1, 1n, (1)n1 1n  1, n1, , (1)n n, 9. a, ln n, n2, , ⬁, , 6. a, n1, ⬁, , 8. a, n2, ⬁, , 10. a, n1, , (1)n1n, 2n 2  1, (1)n1, ln n, (1)n ln(n  1), n2

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774, , Chapter 9 Infinite Sequences and Series, ⬁, , 11. a, n1, ⬁, , 13. a, , (1)n n, 2n, (1)n1 en, pn1, , n0, ⬁, , (2n  1)p, 1, sin, 15. a, 2, n1 1n, sina, , ⬁, , 17. a, n1, , np, b, 2, , 2n 3  1, , ⬁, , p, 19. a (1)n n sina b, n, n1, ⬁, , 21. a, n2, , (1)n ln n, en, , ⬁, , (1)n, , 23. a, n1 1n  1n  1, , ⬁, , 12. a, n1, ⬁, , 14. a, n1, , (1)n1, nen, , 35. Show that the series, 1, 1, 1, 1, 1, 1, 1, 1,     , p n np, 2, 3, 4, 9, 8, 27, 2, 3, , cos np, n, , ⬁, , 16. a (ln n) sin, n2, , converges, and find its sum. Why isn’t the Alternating Series, Test applicable?, , (2n  1)p, 2, , 36. Show that the series, 1, , ⬁, , p, 18. a (1)n cosa b, n, n1, , diverges. Why isn’t the Alternating Series Test applicable?, , ⬁, , (1)n n!, 20. a, nn, n1, ⬁, , 22. a, n2, ⬁, , 24. a, n1, , 37. a. Suppose that 兺 an and 兺 bn are both convergent. Does, it follow that 兺 anbn must be convergent? Justify your, answer., b. Suppose that 兺 an and 兺 bn are both divergent. Does, it follow that 兺 anbn must be divergent? Justify your, answer., ⬁, (1)n, 38. Find all values of s for which a, converges., ns, n1, , (1)n1 1ln n, n, (1)n1, n, , 1n, , ⬁, , (1)n (2n  1), converges., n(n  1), n1, b. Find the sum of the series of part (a)., , In Exercises 25 and 26, find the values of p for which the series, is convergent., ⬁, (1)n, 25. a, p, n2 (ln n), , ⬁, , 26. a (1)n1, n2, , 39. a. Show that a, , (ln n) p, n, , ⬁, , (1)n, converges., n!, n0, b. Denote the sum of the infinite series in part (a) by S., Show that S is irrational., , 40. a. Show that a, , In Exercises 27–30, determine the number of terms sufficient to, obtain the sum of the series accurate to three decimal places., ⬁, (1)n1, 27. a 2, n1 n  1, ⬁, , (2)n3, 29. a, n0 (n  1)!, , ⬁, , 28. a, n1, , (1)n1, , Hint: Use Theorem 2., , 1n, In Exercises 41–44, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, explain why, or give an example to show why it is false., , ⬁, , (1)n1, 30. a, n2 n ln n, , ⬁, 41. If the alternating series 兺n1, (1)n1 an, where an  0,, ⬁, is divergent, then the series 兺n1, an is also divergent., , In Exercises 31–34, find an approximation of the sum of the, series accurate to two decimal places., ⬁, , 31. a, , (1)n, , n1, ⬁, , n3, , (1)n1(n  1), 33. a, 2n, n1, , 9.6, , 1, 1, 1, 1, 1, 1, 1,  ,  , p, , p, 4, 3, 16, 5, 36, 2n  1, (2n)2, , ⬁, 42. Let 兺n1, (1)n1 an be an alternating series, where an  0., ⬁, If lim n→⬁ an  0, then 兺n1, (1)n1 an converges., , ⬁, , (1)n, 32. a, n0 (2n)!, , ⬁, (1)n1 an, where an  0, con43. If the alternating series 兺n1, ⬁, ⬁, verges, then both the series 兺n1, a2n1 and 兺n1, a2n converge., , ⬁, , (1)n1, 34. a, n, n1 n ⴢ 2, , ⬁, (1)n1 an be an alternating series, where an  0., 44. Let 兺n1, ⬁, If an1 an for all n, then 兺n1, (1)n1 an converges., , Absolute Convergence; the Ratio and Root Tests, Absolute Convergence, Up to now, we have considered series whose terms are all positive and series whose, terms alternate between being positive and negative. Now, consider the series, ⬁, , a, n1, , sin 2n, n, , 2, ,  sin 2 , , sin 4, 2, , 2, , , , sin 6, 32, , p

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9.6, , Absolute Convergence; the Ratio and Root Tests, , 775, , With the aid of a calculator you can verify that the first term of this series is positive,, the next two terms are negative, and the next term is positive. Therefore, this series is, neither a series with positive terms nor an alternating series. To study the convergence, of such series, we introduce the notion of absolute convergence., ⬁, an is any series. Then we can form the series, Suppose that 兺n1, ⬁, , p, a 冟 an 冟  冟 a1 冟  冟 a2 冟  冟 a3 冟 , n1, , by taking the absolute value of each term of the given series. Since this series contains, only positive terms, we can use the tests developed in Sections 9.3 and 9.4 to determine its convergence or divergence., , DEFINITION Absolutely Convergent Series, A series 兺 an is absolutely convergent if the series 兺 冟 an 冟 is convergent., Notice that if the terms of the series 兺 an are positive, then 冟 an 冟  an. In this case, absolute convergence is the same as convergence., , EXAMPLE 1 Show that the series, ⬁, , (1)n1, , a, n1, , n2, , 1, , 1, 22, , , , 1, 32, , , , 1, 42, , p, , is absolutely convergent., Solution, , Taking the absolute value of each term of the series, we obtain, ⬁, , a`, , (1)n1, n, , n1, , 2, , ⬁, , ` a, , n1, , 1, n, , 2, , 1, , 1, 2, , 2, , , , 1, 2, , 3, , , , 1, 42, , p, , which is a convergent p-series (p  2). Hence the series is absolutely convergent., , EXAMPLE 2 Show that the alternating harmonic series, ⬁, , (1)n1, 1, 1, 1, 1   p, a, n, 2, 3, 4, n1, is not absolutely convergent., Solution, , Taking the absolute value of each term of the series leads to, ⬁, , a`, , n1, , ⬁, (1)n1, 1, 1, 1, ` a 1  p, n, n, 2, 3, n1, , which is the divergent harmonic series. This shows that the series is not absolutely, convergent., In Example 2 we saw that the alternating harmonic series is not absolutely convergent; but as we proved earlier, it is convergent. Such a series is said to be conditionally convergent.

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776, , Chapter 9 Infinite Sequences and Series, , DEFINITION Conditionally Convergent Series, A series 兺 an is conditionally convergent if it is convergent but not absolutely, convergent., , The following theorem tells us that absolute convergence is, loosely speaking,, stronger than convergence., , THEOREM 1, If a series 兺 an is absolutely convergent, then it is convergent., , PROOF Using an absolute value property, we have, 冟 an 冟, , 冟 an 冟, , an, , Adding 冟 an 冟 to both sides of this inequality yields, 0, , an  冟 an 冟, , 2 冟 an 冟, , If we let bn  an  冟 an 冟, then the last inequality becomes 0 bn 2 冟 an 冟. If 兺 an is, absolutely convergent, then 兺 冟 an 冟 is convergent, which in turn implies, by Theorem 4a, of Section 9.2, that 兺 2 冟 an 冟 is convergent. Therefore, 兺 bn is convergent by the Comparison Test. Finally, since an  bn  冟 an 冟, we see that 兺 an  兺 bn  兺 冟 an 冟 is convergent by Theorem 4b of Section 9.2., As an illustration, the series 兺(1)n1>n 2 of Example 1 is an alternating series that, can be shown to be convergent by the Alternating Series Test. Alternatively, we can, show that the series is absolutely convergent (as was done in Example 1) and conclude, by Theorem 1 that it must be convergent., , EXAMPLE 3 Determine whether the series, ⬁, , a, n1, , sin 2n, n, , 2, ,  sin 2 , , sin 4, 2, , 2, , , , sin 6, 32, , p, , converges or diverges., Solution As was pointed out at the beginning of this section, this series contains both, positive and negative terms, but it is not an alternating series because the first term is, positive, the next two terms are negative, and the next term is positive., Let’s show that the series is absolutely convergent. To do this, we consider the series, ⬁, , a`, , n1, , Since 冟 sin 2n 冟, , sin 2n, n, , 2, , ⬁, , 冟 sin 2n 冟, , n1, , n2, , ` a, , 1 for all n, we see that, 冟 sin 2n 冟, , 1, , n2, , n2, , Now, because 兺 1>n 2 is a convergent p-series, the Comparison Test tells us that, ⬁ 冟, 兺n1, sin 2n 冟>n 2 is convergent. This shows that the given series is absolutely convergent, and we conclude by Theorem 1 that it is convergent.

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778, , Chapter 9 Infinite Sequences and Series, , is a convergent geometric series with 0  r  1 and each term of the series, ⬁, , p, a 冟 aNk 冟  冟 aN1 冟  冟 aN2 冟  冟 aN3 冟 , , (2), , k1, , is less than the corresponding term of the geometric series (1), the Comparison, Test then implies that series (2) is convergent. Since convergence or divergence is, unaffected by the omission of a finite number of terms, we see that the series, ⬁ 冟, 兺n1, an 冟 is also convergent., b. Suppose that, lim `, , n→⬁, , an1, `L1, an, , Let r be any number such that L  r  1. Then there exists an integer N such, that, `, , an1, `r1, an, , whenever n  N. This implies that 冟 an1 冟  冟 an 冟 when n  N. Thus,, lim n→⬁ an 0, and 兺 an is divergent by the Divergence Test., ⬁, ⬁, c. Consider the series 兺n1, 1>n and 兺n1, 1>n 2. For the first series we have, lim `, , n→⬁, , an1, 1, n, `  lim, ⴢ  lim, an, n→⬁ n  1 1, n→⬁, , 1, 1, , 1, n, , 1, , and for the second series we have, lim `, , n→⬁, , an1, 1, n2, `  lim, ⴢ,  lim, an, n→⬁ (n  1) 2, 1, n→⬁, , 1, 1 2, a1  b, n, , 1, , Thus,, lim `, , n→⬁, , an1, `1, an, , for both series. The first series is the divergent harmonic series, whereas the, second series is a convergent p-series with p  2. Thus, if L  1, the series, may converge or diverge, and the Ratio Test is inconclusive., ⬁, , EXAMPLE 4 Determine whether the series a (1)n1, gent, conditionally convergent, or divergent., Solution, , n1, , n2  1, is absolutely conver2n, , We use the Ratio Test with an  (1)n1(n 2  1)>2n. We have, lim `, , n→⬁, , (1)n[(n  1) 2  1], an1, 2n, `  lim `, ⴢ, `, an, n→⬁, 2n1, (1)n1(n 2  1),  lim, , n→⬁, , 1 n 2  2n  2, 1, a, b 1, 2, 2, n2  1, , Therefore, by the Ratio Test, the series is absolutely convergent.

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9.6, , Absolute Convergence; the Ratio and Root Tests, ⬁, , 779, , n!, , EXAMPLE 5 Determine whether the series a n is convergent or divergent., n1 n, Solution, lim `, , n→⬁, , Let an  n!>n n. Then, an1, an1, `  lim, an, n→⬁ an,  lim, , n→⬁, ,  lim, , n→⬁, , (n  1)!, (n  1)n1, ,  lim, , n→⬁, , ⴢ, , nn, n!, , (n  1)n!, nn, ⴢ, (n  1)(n  1)n n!, ,  lim a, n→⬁, , Since an and an1 are positive, , n, n, b, n1, , 1, 1, 1, 1,  lim, ,  1, e, n  1 n n→⬁, 1 n, 1 n, b, a, a1  b, lim a1  b, n, n, n, n→⬁, , Therefore, the series converges, by the Ratio Test., ⬁, , n!, , EXAMPLE 6 Determine whether the series a (1)n n is absolutely convergent,, 3, , conditionally convergent, or divergent., Solution, , n1, , Let an  (1)n n!>3n. Then, lim `, , n→⬁, , (1)n1(n  1)!, an1, 3n, `  lim `, ⴢ, `, an, n→⬁, (1)nn!, 3n1,  lim, , n→⬁, , n1, ⬁, 3, , and we conclude that the given series is divergent by the Ratio Test., Observe that for n  2,, n ⴢ (n  1) ⴢ p ⴢ 3 ⴢ 2 ⴢ 1, n!, 2ⴢ1, 2, , , n , 3, 3ⴢ3ⴢpⴢ3ⴢ3ⴢ3, 3ⴢ3, 9, , Alternative Solution, , 0, , Therefore, lim n→⬁ an  lim n→⬁ (1)nn!>3n does not exist, so the Divergence Test, implies that the series must diverge., , The Root Test, The following test is especially useful when the nth term of a series involves the nth, power. Since the proof is similar to that of the Ratio Test, it will be omitted., , THEOREM 3 The Root Test, ⬁, Let 兺n1, an be a series., n, ⬁, a. If lim n→⬁2冟 an 冟  L  1, then 兺n1, an converges absolutely., n, n, ⬁, b. If lim n→⬁2冟 an 冟  L  1 or lim n→⬁2冟 an 冟  ⬁ , then 兺n1, an diverges., n, c. If lim n→⬁2冟 an 冟  1, the test is inconclusive, and another test should be used.

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780, , Chapter 9 Infinite Sequences and Series, ⬁, , 2n3, , EXAMPLE 7 Determine whether the series a (1)n1, is absolutely con(n  1)n, n1, , vergent, conditionally convergent, or divergent., Solution, , We apply the Root Test with an  (1)n12n3>(n  1)n. We have, n, lim 2冟 an 冟  lim, , n→⬁, , n→⬁, ,  lim, , n→⬁, , n, , B, , ` (1)n1, , 2n3, 2n3 1>n, n `  lim `, n`, (n  1), n→⬁ (n  1), , 213>n, 01, n1, , and conclude that the series is absolutely convergent., , Summary of Tests for Convergence and Divergence of Series, We have developed several ways of determining whether a series is convergent or divergent. Next, we give a summary of the available tests and suggest when it might be, advantageous to use each test., Summary of the Convergence and Divergence Tests for Series, 1. The Divergence Test often settles the question of convergence or divergence of a series 兺 an simply and quickly:, If lim n→⬁ an, , 0, then the series diverges., , 2. If you recognize that the series is, ⬁, a. a geometric series 兺n1, ar n1, then it converges with sum a>(1  r) if, 冟 r 冟  1. If 冟 r 冟  1, the series diverges., b. a telescoping series, then use partial fraction decomposition (if necessary) to find its nth partial sum Sn. Next determine convergence or, divergence by evaluating lim n→⬁ Sn., ⬁, c. a p-series 兺n1, 1>n p, then the series converges if p  1 and diverges if, p 1., Sometimes a little algebraic manipulation might be required to cast the, series into one of these forms. Also, a series might involve a combination, (for example, a sum or difference) of these series., 3. If f(n)  an for n  1, where f is a continuous, positive, decreasing function on [1, ⬁) and readily integrable, then we may use the Integral Test:, ⬁, 兺n1, an converges if 兰1⬁ f(x) dx converges and diverges if 兰1⬁ f(x) dx diverges., , 4. If an is positive and behaves like the nth term of a geometric or p-series, for large values of n, then the Comparison Test or Limit Comparison Test, may be used. The tests and conclusions follow:, a. If an bn for all n and 兺 bn converges, then 兺 an converges., b. If an  bn  0 for all n and 兺 bn  0 diverges, then 兺 an diverges., c. If bn is positive and lim n→⬁ (an>bn)  L  0, then both series converge, or both diverge., The comparison tests can also be used on 兺 冟 an 冟 to test for absolute convergence.

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9.6, , Absolute Convergence; the Ratio and Root Tests, , 781, , ⬁, ⬁, 5. If the series is an alternating series, 兺n1, (1)n an or 兺n1, (1)n1 an, then, the Alternating Series Test should be considered:, , If an  an1 for all n and lim n→⬁ an  0, then the series converges., 6. The Ratio Test is useful if an involves factorials or nth powers. The series, an1, a. converges absolutely if lim `, `  1., n→⬁ an, an1, an1, b. diverges if lim `, `  1 or lim `, `  ⬁., n→⬁ an, n→⬁ an, an1, The test is inconclusive if lim `, `  1., n→⬁ an, 7. The Root Test is useful if an involves nth powers. The series, n, a. converges absolutely if lim n→⬁2冟 an 冟  1., n, n, b. diverges if lim n→⬁2冟 an 冟  1 or lim n→⬁2冟 an 冟  ⬁ ., n, The test is inconclusive if lim n→⬁2冟 an 冟  1., 8. If the series 兺 an involves terms that are both positive and negative but it, is not alternating, then one sometimes can prove convergence of the series, by proving that 兺 冟 an 冟 is convergent., , Rearrangement of Series, A series with a finite number of terms has the same sum regardless of how the terms, of the series are rearranged. The situation gets a little more complicated, however, when, we deal with infinite series. The following example shows that a rearrangement of a, convergent series could result in a series with a different sum!, , EXAMPLE 8 Consider the alternating harmonic series that converges to ln 2 (see, Problem 57 in Exercises 9.8):, 1, , 1, 1, 1, 1, 1, 1, 1,        p  ln 2, 2, 3, 4, 5, 6, 7, 8, , If we rearrange the series so that every positive term is followed by two negative terms,, we obtain, 1, , 1, 1, 1, 1, 1, 1, 1, 1,      , , p, 2, 4, 3, 6, 8, 5, 10, 12, 1, 1, 1, 1, 1, 1, 1, 1,  a1  b   a  b   a  b , p, 2, 4, 3, 6, 8, 5, 10, 12, , , 1, 1, 1, 1, 1, 1,    , , p, 2, 4, 6, 8, 10, 12, , 1, 1, 1, 1, 1, 1, 1,  a1       p b  ln 2, 2, 2, 3, 4, 5, 6, 2, Thus, rearrangement of the alternating harmonic series has a sum that is one half that, of the original series!

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782, , Chapter 9 Infinite Sequences and Series, , You might have noticed that the alternating harmonic series in Example 8 is conditionally convergent. In fact, for such series, Riemann proved the following result:, ⬁, If x is any real number and 兺n1, an is conditionally convergent, then there is a, ⬁, rearrangement of 兺n1 an that converges to x., , A proof of this result can be found in more advanced textbooks., Riemann’s result tells us that for conditionally convergent series, we may not, rearrange their terms, lest we end up with a totally different series, that is, a series with, a different sum. Actually, for conditionally convergent series, one can find rearrangements of the series that diverge to infinity, diverge to minus infinity, or oscillate between, any two prescribed real numbers!, So what kind of convergent series will have rearrangements that converge to the, same sum as the original series? The answer is found in the following result, which, we state without proof:, ⬁, ⬁, ⬁, If 兺n1, an converges absolutely and 兺n1, bn is any rearrangement of 兺n1, an, then, ⬁, ⬁, ⬁, 兺n1 bn converges and 兺n1 an  兺n1 bn., , Finally, since a convergent series with positive terms is absolutely convergent, its, terms can be written in any order, and the resultant series will converge and have the, same sum as the original series., , EXAMPLE 9 Indicate the test(s) that you would use to determine whether the series, converges or diverges. Explain how you arrived at your choice., ⬁, 2n  1, a. a, n1 3n  1, , ⬁, 2, 1, b. a c n , d, 3, n(n,  1), n1, , ⬁, 1, d. a, n1ln, n, n3, , e. a, , ⬁, , g. a (1)n, n1, , ⬁, n3, , 1n, n 1, 2, , ⬁, , ln n, n, , ⬁, 1 e, c. a a b, n1 n, , f. a, , 2, , n1, , ⬁, n, h. a n, n1 2, , ⬁, , i. a, , 2n 3  2, n  3n 2  1, 4, , sin n, , n1 2n, , 3, , 1, , Solution, a. Since, lim an  lim, , n→⬁, , n→⬁, , 2n  1, 2, , 3n  1, 3, , 0, , we use the Divergence Test., b. The series is the difference of a geometric series and a telescoping series, so we, use the properties of these series to determine convergence., 1 e, 1, c. Here, an  a b  e is a p-series, so we use the properties of a p-series to, n, n, study its convergence., 1, d. The function f(x) , is continuous, positive, and decreasing on [3, ⬁) and, x1ln x, is integrable, so we choose the Integral Test., e. Here,, ln n, 1n, 1, an  2  2  3>2  bn, n, n, n, and we use the Comparison Test with the test series 兺 bn.

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9.6, , f. an , , (n 3  2)1>2, n 4  3n 2  1, , Absolute Convergence; the Ratio and Root Tests, , 783, , is positive and behaves like, bn , , (n 3)1>2, n, , 4, , , , n 3>2, n4, , , , 1, n 5>2, , for large values of n, so we use the Limit Comparison Test with test series, ⬁, 兺n1, 1>n 5>2., g. This is an alternating series, and we use the Alternating Series Test., n, n 1>n n, h. Here, an  n  a, b involves the nth power, so the Root Test is a candidate., 2, 2, n, 1n, 1, n, In fact, here lim 2冟 an 冟  lim,   1 and the series converges., n→⬁, n→⬁ 2, 2, i. The series involves both positive and negative terms and is not an alternating, series, so we use the test for absolute convergence., , 9.6, , CONCEPT QUESTIONS, 2. a. State the Ratio Test and the Root Test., b. Give an example of a convergent series and an example, of a divergent series for which the Ratio Test is inconclusive., c. Give an example of a convergent series and an example, of a divergent series for which the Root Test is inconclusive., , 1. a. What is an absolutely convergent series? Give an, example., b. What is a conditionally convergent series? Give an, example., , 9.6, , EXERCISES, , In Exercises 1–34, determine whether the series is convergent,, absolutely convergent, conditionally convergent, or divergent., ⬁, , 1. a, , (1)n1, 1n, , n1, ⬁, , 3. a, , (2), , n1, , n1, , n2, , ⬁, , (1)n1, 5. a, n1 n  1, ⬁, , n 2, , (1) n, 7. a 2, n1 n  3, ⬁, (1)n, 9. a, n2 n ln n, ⬁, , n!, 11. a n, n1 e, ⬁, 1, 13. a (1)n1 sina b, n, n1, , ⬁, (1)n, 2. a, n1 n1n, ⬁, , 4. a, n1, ⬁, , (2), n!, , n, , 8. a, n1, , 12. a, n1, ⬁, , 14. a, n1, , 18. a, , n2, , (1)nn, (1), , n1, , n, , 22n  1, , ⬁, , cos(n  1), n1n, n1, , n, , tan, 2, , (1)n ln n, 2n, , np, b, ⬁ sina, 4, 21. a, 2, n2 n(ln n), , 2, , (1), , ⬁, (2)nn, 17. a, n1, n1 (n  1)3, ⬁, , ⬁, (1)n, 10. a, n3 n1ln n, ⬁, , ⬁, (5)n1, 16. a 2 n, n1 n ⴢ 3, , 19. a, , 6. a 2, n1 n  1, ⬁, , ⬁, 2n, 15. a, n1 n! n, , 1, , n, , ⬁, , (1)n1 ln n, n2  1, , n2, ⬁, , 20. a, n0, , ⬁, , 22. a, n1, , cos np, n!, (1)n1n 5, en, , (1)n1n n, 23. a, n!, n1, , ⬁, ln n n, 24. a a, b, n, n2, , ⬁, (1)n, 25. a, n, n2 (ln n), , ⬁, n, n, b, 26. a a, n1 2n  1, , ⬁, 1, 27. a (1)n tana b, n, , ⬁, p, 28. a (1)n n sina b, n, , n1, ⬁, , (n)n, , 29. a, 1, n]n, n1 [(n  1)tan, , V Videos for selected exercises are available online at www.academic.cengage.com/login., , n1, , 30. a 1 1n  1 2 n, ⬁, , n2, , n

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784, , Chapter 9 Infinite Sequences and Series, ⬁, np n, where 0  p  1, is con43. a. Show that the series 兺n1, vergent., p, b. Show that its sum is S , ., (1  p)2, , ⬁, 3 ⴢ 5 ⴢ 7 ⴢ p ⴢ (2n  1), 31. a (1)n1, 1 ⴢ 4 ⴢ 7 ⴢ p ⴢ (3n  2), n1, ⬁, 2n, 32. a (1)n, p, 3 ⴢ 5 ⴢ 7 ⴢ ⴢ (2n  1), n1, ⬁, 4 ⴢ 7 ⴢ 10 ⴢ p ⴢ (3n  1), 33. a, 4n(n  1)!, , Hint: Find an expression for Sn  pSn., , 44. Average Number of Coin Tosses An unbiased coin is tossed until, the coin lands heads and the number of throws in the experiment is recorded. As more and more experiments are performed, the average number of tosses obtained from these, ⬁, experiments approaches 兺n1, n 1 12 2 n. Use the result of Exercise 43 to find this number., , n1, , ⬁, (n!)2, 34. a, n1 (3n)!, ⬁, xn, 35. Find all values of x for which the series a, (a) conn1 n, verges absolutely and (b) converges conditionally., , ⬁, 冟 an 冟 converges, then so does, 45. Show that if 兺n1, ⬁, 兺n2 冟 an  an1 冟., , 36. Show that the Ratio Test is inconclusive for the p-series., , ⬁, 46. Show that if 兺n1, an is absolutely convergent, then, ⬁, ⬁, 冟 兺n1, 冟, an, 兺n1 冟 an 冟., , 37. Show that the Root Test is inconclusive for the p-series., 38. a. Show that if 兺 an converges absolutely, then 兺 a 2n converges., b. Show that the converse of the result in part (a) is false by, finding a series 兺 an for which 兺 a 2n converges, but 兺 冟 an 冟, diverges., 39. Show that if 兺 an diverges, then 兺 冟 an 冟 diverges., 40. Show that if 兺 an converges absolutely, then 兺 an, , 兺 冟 an 冟., , 41. Suppose that 兺 a 2n and 兺 b 2n are convergent. Show that, 兺 anbn is absolutely convergent., Hint: Show that 2 冟 ab 冟, , In Exercises 47–50, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, explain why, or give an example to show why it is false., ⬁, ⬁, 47. If 兺n1, an and 兺n1, bn converge absolutely, then, ⬁, 兺n1(an  bn) converges absolutely., ⬁, 48. If an  0 for n  1 and 兺n1, an converges, then, ⬁, 兺n1, (1)nan converges., ⬁, ⬁, ⬁, 49. If 兺n1, 2a 2n  b 2n converges, then 兺n1, an and 兺n1, bn converge absolutely., , a 2  b 2 by looking at (a  b)2 and, , ⬁, , 0 for any n  1 and a an converges absolutely,, n1, ⬁, 1, then a, diverges., 冟 冟, n1 an, , (a  b) ., 2, , 50. If an, , 42. Prove that lim, , n→⬁, , 2nn!,  0., nn, ⬁, , 2nn!, , Hint: Show that a n is convergent., n1 n, , 9.7, , Power Series, Power Series, Until now, we have dealt with series with constant terms. In this section we will study, infinite series of the form, ⬁, , n, 2, 3, n, p, p, a anx  a0  a1x  a2x  a3x   anx , n0, , where x is a variable. More generally, we will consider series of the form, ⬁, , n, 2, 3, n, p, p, a an(x  c)  a0  a1 (x  c)  a2 (x  c)  a3 (x  c)   an (x  c) , n0, ⬁, from which 兺n0, anx n may be obtained as a special case by putting c  0. We may, view such series as generalizations of the notion of a polynomial to an infinite series., Examples of power series are, ⬁, , n, 2, 3, p, ax 1xx x , n0

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9.7, , Power Series, , 785, , ⬁, , (1)nx n, x2, x3 p, , 1, , x, , , , a, n!, 2!, 3!, n0, and, ⬁, , a, , (1)n 1 x  p4 2 2n1, (2n  1)!, , n0, ,  ax , , 1 x  p4 2 3 1 x  p4 2 5, p, b, , p, 4, 3!, 5!, , Observe that if we truncate each of these series, we obtain a polynomial., , DEFINITION Power Series, Let x be a variable. A power series in x is a series of the form, ⬁, , n, 2, 3, n, p, p, a anx  a0  a1x  a2x  a3x   anx , n0, , where the an’s are constants and are called the coefficients of the series. More, generally, a power series in (x ⴚ c), where c is a constant, is a series of the, form, ⬁, , n, 2, a an(x  c)  a0  a1(x  c)  a2(x  c),  a3 (x  c)3  p  an(x  c)n  p, , n0, , Notes, 1. A power series in (x  c) is also called a power series centered at c or a power, series about c. Thus, a power series in x is just a series centered at the origin., 2. To simplify the notation used for a power series, we have adopted the convention, that (x  c)0  1, even when x  c., We can view a power series as a function f defined by the rule, ⬁, , f(x)  a an(x  c)n, n0, , The domain of f is the set of all x for which the power series converges, and the range, of f comprises the sums of the series obtained by allowing x to take on all values in, the domain of f. If a function f is defined in this manner, we say that f is represented, ⬁, an(x  c)n., by the power series 兺n0, , EXAMPLE 1 As an example, consider the power series, ⬁, , n, 2, 3, p  xn  p, ax 1xx x , , (1), , n0, , Recognizing that this is a geometric series with common ratio x, we see that it converges for 1  x  1. Thus, the power series (1) is a rule for a function f with interval (1, 1) as its domain; that is,, ⬁, , f(x)  a x n  1  x  x 2  x 3  p  x n  p, n0, , There is a simple formula for the sum of the geometric series (1), namely, 1>(1  x),

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786, , Chapter 9 Infinite Sequences and Series, , y, , y  f (x), , and we see that the function represented by the series is the function, f(x) , , 2, , 2, , (, 1, , 0, , ), 1, , 2, , x, , 1, 1x, , 1  x  1, , Even though the domain of the function t(x)  1>(1  x) is the set of all real numbers except x  1, the power series (1) represents the function t(x)  1>(1  x) only, in the interval of convergence (1, 1) of the series. (See Figure 1.) Observe that the, ⬁, nth partial sum Sn(x)  1  x  x 2  p  x n of 兺n0, x n approximates t(x) better and, better as n increases for 1  x  1. But outside this interval, Sn (x) diverges from, t(x). (See Figure 2.), , FIGURE 1, ⬁, The function f(x)  兺n0, xn, represents the function, 1, t(x) , on (1, 1) only., 1x, , y, , y  g(x), , S2 S3, , 4, , S1, , 2, S0, 2, , 1, , 0, , 1, , 2, , 3 x, , 2, , FIGURE 2, Observe that Sn(x)  兺nk0 x k, approximates t(x) better and, better as n → ⬁ for 1  x  1., , 4, , Example 1 reveals one shortcoming in representing a function by a power series., But as we will see later on, the advantages far outweigh the disadvantages., , Interval of Convergence, How do we find the domain of a function represented by a power series? Suppose that, f is the function represented by the power series, ⬁, , f(x)  a an(x  c)n  a0  a1 (x  c)  a2 (x  c)2, n0,  a3(x  c)3  p  an (x  c)n  p, , (2), , Since f(c)  a0, we see that the domain of f always contains at least one number (the, center of the power series) and is therefore nonempty. The following theorem, which, we state without proof, tells us that the domain of a power series is always an interval, with x  c as its center. In the extreme cases the domain consists of the infinite interval (⬁, ⬁) or just the point x  c, which may be regarded as a degenerate interval., , THEOREM 1 Convergence of a Power Series, ⬁, Given a power series 兺n0, an (x  c)n, exactly one of the following is true:, , a. The series converges only at x  c., b. The series converges for all x., c. There is a number R  0 such that the series converges for 冟 x  c 冟  R, and diverges for 冟 x  c 冟  R., , A proof of Theorem 1 is given in Appendix B.

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9.7, , Power Series, , 787, , The number R referred to in Theorem 1 is called the radius of convergence of the, power series. The radius of convergence is R  0 in case (a) and R  ⬁ in case (b)., The set of all values for which the power series converges is called the interval of convergence of the power series. Thus, Theorem 1 tells us that the interval of convergence, of a power series centered at c is (a) just the single point c, (b) the interval (⬁, ⬁),, or (c) the interval (c  R, c  R). (See Figure 3.) But in the last case, Theorem 1 does, not tell us whether the endpoints x  c  R and x  c  R are included in the interval of convergence. To determine whether they are included, we simply replace x in, the power series (2) by c  R and c  R in succession and use a convergence test on, the resultant series., , R, Series converges, , Series, diverges, , FIGURE 3, ⬁, The power series 兺n0, an(x  c)n, converges for 冟 x  c 冟  R, and diverges for 冟 x  c 冟  R., , (, cR, , c, , Series, diverges, ), cR, , x, , EXAMPLE 2 Find the radius of convergence and the interval of convergence of, ⬁, 兺n0, n! x n., ⬁, Solution We can think of the given series as 兺n0, u n, where u n  n! x n. Applying the, Ratio Test, we have, , lim `, , n→⬁, , (n  1)! x n1, u n1, `  lim `, `  lim (n  1) 冟 x 冟  ⬁, un, n→⬁, n! x n, n→⬁, , whenever x 0, and we conclude that the series diverges whenever x 0. Therefore,, the series converges only when x  0, and its radius of convergence is accordingly, R  0., , EXAMPLE 3 Find the radius of convergence and the interval of convergence of, ⬁, , a, n0, , Solution, , (1)n x 2n, (2n)!, , Let, un , , (1)n x 2n, (2n)!, , Then, lim `, , n→⬁, , (1)n1 x 2n2, (2n)!, u n1, `  lim `, ⴢ, `, un, n→⬁, (2n  2)!, (1)n x 2n,  lim, , n→⬁, , x2, 01, (2n  1)(2n  2), , for each fixed value of x, so by the Ratio Test, the given series converges for all values of x. Therefore, the radius of convergence of the series is R  ⬁ , and its interval, of convergence is (⬁, ⬁).

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788, , Chapter 9 Infinite Sequences and Series, , EXAMPLE 4 Find the radius of convergence and the interval of convergence of, ⬁, , xn, ., a, n1 n, Solution, , Let u n  x n>n. Then, lim `, , n→⬁, , u n1, x n1, n, n, `  lim `, ⴢ n `  lim a, b冟 x 冟  冟 x 冟, un, n→⬁ n  1 x, n→⬁ n  1, , By the Ratio Test, the series converges if 冟 x 冟  1, that is, if 1  x  1. Therefore,, the radius of convergence of the series is R  1. To determine the interval of convergence of the power series, we need to examine the behavior of the series at the endpoints x  1 and x  1. Now, if x  1, the series becomes, ⬁, , a, n1, , [, 1, , 0, , ), 1, , FIGURE 4, The interval of convergence, ⬁, of 兺n1, x n>n is the interval [1, 1), with center c  0 and radius R  1., , x, , (1)n, n, , which is the convergent alternating harmonic series, and we see that x  1 is in the, interval of convergence of the power series. If x  1, we obtain the harmonic series, ⬁, 兺n1, 1>n, which is divergent, so x  1 is not in the interval of convergence. We conclude that the interval of convergence of the given power series is [1, 1), as shown, in Figure 4., , EXAMPLE 5 Find the radius of convergence and the interval of convergence of, ⬁, , a, n1, , (x  2)n, n 2 ⴢ 3n, , Solution, , ., , Letting, un , , (x  2)n, n 2 ⴢ 3n, , we have, lim `, , n→⬁, , (x  2)n1, u n1, n 2 ⴢ 3n, `  lim `, ⴢ, `, un, n→⬁ (n  1) 23n1, (x  2)n,  lim a, n→⬁, , 2 冟, 冟x  2冟, n, x  2冟, b, , n1, 3, 3, , By the Ratio Test, the series converges if 冟 x  2 冟>3  1 or 冟 x  2 冟  3. The last, inequality tells us that the radius of convergence of the given series is R  3 and that, the power series converges for x in the interval (1, 5)., Next, we check the endpoints x  1 and x  5. If x  1, the power series, becomes, ⬁, , (3)n, , a, n1, , n 2 ⴢ 3n, , ⬁, , (1)n, , n1, , n2, ,  a, , which is a convergent alternating series. Therefore, x  1 is in the interval of convergence. Next, if x  5, we obtain, ⬁, , a, n1, , ⬁, , 3n, n ⴢ3, 2, , n, ,  a, n1, , 1, n2

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9.7, , [, 1, , 0, , 2, , ], 5, , FIGURE 5, The interval of convergence of, ⬁, (x  2)n, a n 2 ⴢ 3n is the interval [1, 5], , x, , Power Series, , 789, , which is a convergent p-series. Therefore, x  5 is also in the interval of convergence., We conclude, accordingly, that the interval of convergence of the given power series, is [1, 5], as shown in Figure 5., , n1, , with center c  2 and radius R  3., , EXAMPLE 6 Find the radius of convergence and the interval of convergence of, ⬁, , (1)n2nx n, ., a, n0 1n  1, Solution, , Let, un , , (1)n2nx n, 1n  1, , Then, , Historical Biography, , lim `, , n→⬁, , (1)n12n1x n1, u n1, 1n  1, `  lim `, ⴢ, `, un, n→⬁, (1)n2nx n, 1n  2, , Mary Evans Picture Library/, The Image Works, ,  lim 2, n→⬁, , FRIEDRICH WILHELM BESSEL, (1784–1846), The first person to use the term light-years, as a way to express extreme distances,, Friedrich Bessel astounded other astronomers when he proved that one of the, nearest stars to the earth, 61 Cygni, was, about 10 light-years (more that 60 trillion, miles) away. He was able to apply his, method of computation to compile a catalog of the positions of 50,000 stars. This, was no small task for a mathematician and, astronomer who did not have a university, education. Born in Minden, Germany, on, July 22, 1784, Bessel began to study navigation, geography, and foreign languages, after he decided to enter foreign trade., From the study of navigation, his interest, of astronomy blossomed, and his mathematical and analytical talents quickly surfaced. He made many discoveries in astronomy, and his mathematical discoveries, were often derived from his work in that, field. The so-called Bessel functions (see, Exercise 32 in this section) were studied, extensively by Bessel and are still important in mathematics, with important applications in areas such as geology, physics,, and engineering., , 1  (1>n), n1, 冟 x 冟  2 冟 x 冟 lim,  2冟x冟, Bn  2, n→⬁ B 1  (2>n), , By the Ratio Test, the series converges if 2 冟 x 冟  1 or 冟 x 冟  12. The last inequality, tells us that the radius of convergence of the power series is R  12, and the series converges in the interval 1 12, 12 2 ., Next, we check the endpoints x  12 and x  12. If x  12, the power series, becomes, ⬁, , a, n0, , (1)n2n 1 12 2 n, 1n  1, , ⬁, 1,  a, n0 1n  1, , which can be shown to be divergent by the Limit Comparison Test. (Compare it with, ⬁, the p-series 兺n1, 1>n 1>2.) Next, if x  12, we have, ⬁, , a, n0, , (1)n2n 1 12 2 n, 1n  1, , ⬁, (1)n,  a, n0 1n  1, , which converges, by the Alternating Series Test. Therefore, the interval of convergence, of the power series is 1 12, 12 D ., , Differentiation and Integration of Power Series, Suppose that f is a function represented by a power series centered at c, that is,, ⬁, , f(x)  a an(x  c)n, n0, , where x lies in the interval of convergence of the series (domain of f). The following, question arises naturally: Can we differentiate and integrate f, and if so, what are the, series representations of the derivative and integral of f ? The next theorem answers this, question in the affirmative and tells us that the series representations of the derivative, and integral of f are found by differentiating and integrating the power series representation of f term by term. (We omit its proof.)

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790, , Chapter 9 Infinite Sequences and Series, , THEOREM 2 Differentiation and Integration of Power Series, ⬁, Suppose that the power series 兺n0, an(x  c)n has a radius of convergence, R  0. Then the function f defined by, ⬁, , f(x)  a an (x  c)n  a0  a1(x  c)  a2(x  c)2  a3(x  c)3  p, n0, , for all x in (c  R, c  R) is both differentiable and integrable on, (c  R, c  R). Moreover, the derivative of f and the indefinite integral of f are, ⬁, , a. f ¿(x)  a1  2a2(x  c)  3a3(x  c)2  p  a nan (x  c)n1, n1, , b., , 冮 f(x) dx  C  a (x  c)  a, 0, , ⬁, ,  a an, n0, , 1, , (x  c), (x  c)3 p,  a2, , 2, 3, 2, , (x  c)n1, C, n1, , Notes, 1. The series in parts (a) and (b) of Theorem 2 have the same radius of convergence,, ⬁, R, as the series 兺n0, an(x  c)n. But the interval of convergence may change., More specifically, you may lose convergence at the endpoints when you differentiate (Exercise 38) and gain convergence there when you integrate (Example 9)., 2. Theorem 2 implies that a function that is represented by a power series in an, interval (c  R, c  R) is continuous on that interval. This follows from Theorem 1 in Section 2.1., , EXAMPLE 7 Find a power series representation for 1>(1  x)2 on (1, 1) by differ-, , entiating a power series representation of f(x)  1>(1  x)., Solution, , Recalling that 1>(1  x) is the sum of a geometric series, we have, f(x) , , ⬁, 1,  1  x  x2  x3  p  a xn, 1x, n0, , 冟x冟  1, , Differentiating both sides of this equation with respect to x and using Theorem 2, we, obtain, f ¿(x) , , 1, (1  x)2, , ⬁, ,  1  2x  3x 2  p  a nx n1, n1, , EXAMPLE 8 Find a power series representation for ln(1  x) on (1, 1)., Solution, , We start with the equation, ⬁, 1,  1  x  x2  x3  p  a xn, 1x, n0, , 冟x冟  1, , Integrating both sides of this equation with respect to x and using Theorem 2, we obtain, , 冮 1  x dx  冮 (1  x  x, 1, , 2, ,  x 3  p) dx

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9.7, , Power Series, , 791, , or, ln(1  x)  x , , 1 2 1 3 p, x  x  C, 2, 3, , To determine the value of C, we set x  0 in this equation to obtain ln 1  0  C., Using this value of C, we see that, ln(1  x)  x , , ⬁, 1 2 1 3 p, xn, x  x  a, 2, 3, n1 n, , 冟x冟  1, , EXAMPLE 9 Find a power series representation for tan1 x by integrating a power, series representation of f(x)  1>(1  x 2)., Solution Observe that we can obtain a power series representation of f by replacing, x with x 2 in the equation, 1,  1  x  x2  p, 1x, , 冟x冟  1, , Thus,, 1, 1x, , 2, , , , 1, ,  1  (x 2)  (x 2)2  (x 2)3  p, , 1  (x 2), , ⬁, ,  1  x 2  x 4  x 6  p  a (1)n x 2n, n0, , Since the geometric series converges for 冟 x 冟  1, we see that this series converges for, 冟 x 2 冟  1, that is, x 2  1 or 冟 x 冟  1. Finally, integrating this equation, we have, by, Theorem 2,, tan1 x , , 冮1  x, 1, , 2, , Cx, , dx , , 冮 (1  x, , 2, ,  x 4  x 6  p) dx, , x3, x5, x7 p, , , , 3, 5, 7, , To find C, we use the condition tan1 0  0 to obtain 0  C. Therefore,, tan1 x  x , , ⬁, x3, x5, x7 p, x 2n1, , ,   a (1)n, 3, 5, 7, 2n  1, n0, , We leave it for you to show that the interval of convergence of the series is [1, 1]., , 9.7, 1. a., b., 2. a., b., c., , CONCEPT QUESTIONS, , Define a power series in x., Define a power series in (x  c)., What is the radius of convergence of a power series?, What is the interval of convergence of a power series?, How do you find the radius and the interval of convergence of a power series?, , ⬁, 3. Suppose that 兺n0, an x n has radius of convergence 2. What, can you say about the convergence or divergence of, ⬁, 兺n0, an 1 32 2 n?, ⬁, 4. Suppose that 兺n0, an(x  2)n diverges for x  0. What can, ⬁, you say about the convergence or divergence of 兺n0, an5n?, ⬁, n, What about 兺n0 an2 ?

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792, , Chapter 9 Infinite Sequences and Series, , 9.7, , EXERCISES, b. Evaluate lim n→⬁ Rn(x) for each fixed x in the interval, (1, 1). What happens to lim n→⬁ Rn(x) for 冟 x 冟  1?, c. Plot the graphs of Rn (x) for n  1, 2, 3, p , 5, and 20, using the viewing window [2, 2] [10, 5]., , In Exercises 1–30, find the radius of convergence and the, interval of convergence of the power series., ⬁, xn, 1. a, n0 n  1, , ⬁, , 2. a (1)n1 nx n, n1, , ⬁, , ⬁, , 5. a, n0, , ⬁, , (2x)n, n!, , n1, ⬁, xn, 9. a, n2 ln n, , n2, ⬁, , 12. a, n0, , (1) (x  3), n, , ⬁, , 19. a, n0, , ⬁, , n 1, 2, , 34. If a is a constant, find the radius and the interval of conver⬁, a n (x  c)n., gence of the power series 兺n0, , ⬁, n(x  2)n, 18. a 2, n, n0 (n  1)2, , 35. If the radius of convergence of the power series 兺 anx n is R,, what is the radius of convergence of the power series 兺 anx 2n?, , ⬁, , (1)n(x  2)2n1, (2n  1)!, , 20. a, n0, ⬁, , 2n(x  2)n, 21. a, nn, n1, (1) (3x  5), 23. a, n ln n, n2, , is called the Bessel function of order 1. What is its domain?, , n(2x  1)n, 16. a, 2n, n1, , ⬁, , ⬁, , ⬁, (1)n x 2n1, J1(x)  a, 2n1, n0 n!(n  1)! 2, , n1, , (1)n n(x  1)n, , n, , 33. A Bessel Function The function J1 defined by, , 14. a 1n(2x  3)n, , 1n, , (1)n1(x  2)n, 15. a, n ⴢ 3n, n1, , n0, , (1)n n! x n, 2n, , ⬁, , n, , ⬁, , ⬁, , is called the Bessel function of order 0., a. What is the domain of J0?, b. Plot the graph of J0 in the viewing window, [10, 10] [0.5, 1.2], and plot the graphs, of Sn (x) for n  0, 1, 2, 3, and 4 in the viewing, window [8, 8] [2, 2]., , ⬁, , en x n, 11. a, n, n1, , 17. a, , n, , 10. a (x ln n)n, , ⬁, , n1, , ⬁, , ⬁, (1)n x 2n, J0(x)  a 2n, 2, n0 2 (n!), , n! x, 8. a, n0 (2n)!, , 7. a (nx)n, , 13. a, , n1, , (1)n x n, n ⴢ 3n, , 6. a, , ⬁, , ⬁, , 32. A Bessel Function The function J0 defined by, , ⬁, xn, 4. a 2, n1 n, , xn, 3. a, n1 1n, , 22. a, , (1)n(3x  2)2n, (2n)!, (3x  1)n, n3  n, , n1, ⬁, , (1)n(x  2)n, 24. a, (ln n) n, n2, , n, , ⬁, xn, 25. a, 2, n2 n(ln n), , ⬁, , 26. a, n1, , n n(3x  5)n, (2n)!, , ⬁, , 2 ⴢ 4 ⴢ 6 ⴢ p ⴢ 2n, 2n1, 27. a, p ⴢ (2n  1) x, n1 3 ⴢ 5 ⴢ 7 ⴢ, ⬁, , (1)n n! (x  1)n, 28. a, p ⴢ (2n  1), n1 1 ⴢ 3 ⴢ 5 ⴢ, ⬁, (1)n 2n n! x n, 29. a, p ⴢ (3n  2), n1 5 ⴢ 8 ⴢ 11 ⴢ, n, ⬁, (1) 2 ⴢ 4 ⴢ 6 ⴢ p ⴢ 2n(x  p)n, 30. a, n!, n1, , ⬁, 兺n0, , n, , x and the (sum) function, 31. Consider the series, f(x)  1>(1  x) represented by the series for 1  x  1., a. Find the remainder Rn (x)  f(x)  Sn(x), where, ⬁, Sn(x)  兺nk0 x k is the nth partial sum of 兺n0, xn, and x is fixed., , 36. Suppose that lim n→⬁ 冟 an1>an 冟  L and L 0. Show that, the radius of convergence of the power series 兺 anx n is 1>L., n, 37. Suppose that lim n→⬁2冟 an 冟  L and L 0. What is the, radius of convergence of the power series 兺 anx n?, ⬁, , 38. Let f(x)  a, , (x  2)n, , . Show that the domain of f is, n 23n, [1, 5] but the domain of f ¿ is [1, 5)., n1, , ⬁, xn, 39. Let f(x)  a 2 . Find f ¿(x) and f ⬙(x). What are the intern1 n, vals of convergence of f, f ¿, and f ⬙?, ⬁, , 40. Show that the series a, n1, 3, , sin(n 3x), n2, , converges for all values, , ⬁, d sin(n x), c, d diverges for all values of x. Does, of x, but a, dx, n2, n1, this contradict Theorem 2? Explain your answer., ⬁, nx n1, 冟 x 冟  1., 41. Find the sum of the series 兺n1, ⬁, Hint: Differentiate the geometric series 兺n0, x n., , ⬁, 42. a. Find the sum of the series 兺n1, nx n, 冟 x 冟  1., , Hint: See the hint for Exercise 41., , ⬁, n, b. Use the result of part (a) to find the sum of a n ., 2, n1, , 43. Suppose that the interval of convergence of the series, ⬁, an(x  c)n is (c  R, c  R]. Prove that the series, 兺n0, is conditionally convergent at c  R., , V Videos for selected exercises are available online at www.academic.cengage.com/login.

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9.8, ⬁, 44. Suppose that the series 兺n0, an(x  c)n is absolutely convergent at one endpoint of its interval of convergence. Prove, that the series is also absolutely convergent at the other endpoint., , 45. a. Find a power series representation for 1>(1  t 2)., b. Use the result of part (a) to find a power series representation of tanh1 x using the relationship, x, , tanh1 x , , 冮 1t, 1, , 0, , 2, , dt, , 46. Use the result of Example 8, ⬁, , ln(1  x)   a, n1, , D, , m 2t, m√0, k√0,  2 (sin a)lna1 , b, k, mt sin a, k, , where k is the constant of proportionality and t is the constant of acceleration due to gravity., a. Show that D can be written as, 2, √30, √40, 1, 1, 1 √0, , k, k2  p, 2, 2, 2 t sin a, 3 m(t sin a), 4 m (t sin a) 3, , Hint: Use the result of Example 8., , b. Use the result of part (a) to show that in the absence of, air resistance the object reaches a maximum distance of, √20>(2t sin a) up the incline., , xn, n, , to obtain an approximation of ln 1.2 accurate to five decimal, places., Hint: Use Theorem 2 in Section 9.5., , 47. Use the result of Example 9,, ⬁, x 2n1, tan1 x  a (1)n, 2n  1, n0, , to obtain an approximation of p accurate to five decimal, places., Hint: Use Theorem 2 in Section 9.5., , 48. Motion Along an Inclined Plane An object of mass m is thrown, up an inclined plane that makes an angle of a with the horizontal. If air resistance proportional to the instantaneous, , 9.8, , 793, , velocity is taken into consideration, then the object reaches a, maximum distance up the incline given by, , D, , What is the radius of convergence of the series?, , Taylor and Maclaurin Series, , In Exercises 49–52, determine whether the statement is true or, false. If it is true, explain why it is true. If it is false, explain why, or give an example to show why it is false., ⬁, 49. If the power series 兺n0, anx n converges for x  3, then it, converges for x  2., ⬁, 50. If the power series 兺n0, anx n converges for x in (1, 1),, ⬁, then f(x)  兺n0, anx n is continuous on (1, 1)., ⬁, 51. If the interval of convergence of 兺n0, anx n is [2, 2), then, ⬁, the interval of convergence of 兺n0 an(x  3)n is [1, 5)., ⬁, 52. If the radius of convergence of 兺n0, anx n is R  0, then the, 1 ⬁ an, radius of convergence of the power series in , a n ,, x n0 x, 1, is ., R, , Taylor and Maclaurin Series, In Section 9.7 we saw that every power series represents a function whose domain is, precisely the interval of convergence of the series. We also touched upon the converse, problem: Given a function f defined on an interval containing a point c, is there a power, series centered at c that represents f, and if so, how do we find it? There, we were able, to look only at functions whose power series representations are obtained by manipulating the geometric series., We now look at the general problem of finding power series representations for, functions. The problem centers on finding the answers to two questions:, 1. What form does the power series representation of the function f take? (In other, words, what does an look like?), 2. What conditions will guarantee that such a power series will represent f ?, We will consider the first question here and leave the second for Section 9.9., , Taylor and Maclaurin Series, Suppose that f is a function that can be represented by a power series that is centered, at c and has a radius of convergence R  0. If 冟 x  c 冟  R, we have, f(x)  a0  a1 (x  c)  a2(x  c)2  a3(x  c)3  a4(x  c)4  p  an (x  c)n  p

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794, , Chapter 9 Infinite Sequences and Series, , Applying Theorem 2 of Section 9.7 repeatedly, we obtain, f ¿(x)  a1  2a2(x  c)  3a3(x  c)2  4a4(x  c)3  p  nan (x  c)n1  p, f ⬙(x)  2a2  3 ⴢ 2a3(x  c)  4 ⴢ 3a4(x  c)2  p  n(n  1)an (x  c)n2  p, f ‡(x)  3 ⴢ 2a3  4 ⴢ 3 ⴢ 2a4(x  c)  p  n(n  1)(n  2)an(x  c)n3  p, o, f, , (n), , (x)  n(n  1)(n  2)(n  3) ⴢ p ⴢ 2an  p, o, , Each of these series is valid for x satisfying 冟 x  c 冟  R. Substituting x  c in each, of the above expressions, we obtain, f(c)  a0,, , f ¿(c)  a1,, , f ‡(c)  3! a3,, , Historical Biography, , p,, , f ⬙(c)  2a2,, (c)  n! an,, , p, , from which we find, a0  f(c),, , Topham/The Image Works, , f, , (n), , a3 , , a1  f ¿(c),, , f ‡(c), ,, 3!, , an , , p,, , a2 , , f ⬙(c), ,, 2!, , f (n) (c), ,, n!, , p, , We have proved that if f has a power series representation, then the series must, have the form given in the following theorem., BROOK TAYLOR, (1685–1731), , Born to a family of minor nobility, Brook, Taylor was tutored at home until he was, admitted to St. John’s College in Cambridge, England, in 1703. There, he completed a bachelor of law degree in 1709,, but he had already written his first paper, in mathematics. In 1712 he was elected to, the Royal Society, where he was appointed, to a committee to decide whether Newton, (page 179) or Leibniz (page 157) had first, invented the calculus. Taylor wrote numerous articles during his time with the Royal, Society, among them a work entitled, Methodus incrementorum directa et, inversa. This work contained the theorem, for which Taylor is most remembered: the, so-called Taylor series. Other mathematicians had developed such series for specific functions, but none before Taylor had, given a general series expansion for functions of a single variable. Taylor’s mathematical work was of great depth, but his, writing style was concise and hard to follow. He often left much for the reader to, work out, and this delayed the acknowledgment of the majority of his accomplishments until after his death., , THEOREM 1 Taylor Series of f at c, If f has a power series representation at c, that is, if, ⬁, , f(x)  a an (x  c)n, , 冟x  c冟  R, , n0, , then f (n) (c) exists for every positive integer n and, an , , f (n) (c), n!, , Thus,, ⬁, , f(x)  a, n0, , f (n) (c), (x  c)n, n!, , f ⬙(c), f ‡(c),  f(c)  f ¿(c)(x  c) , (x  c)2 , (x  c)3  p, 2!, 3!, , (1), , A series of this form is called the Taylor series of the function f at c after the, English mathematician Brook Taylor (1685–1731)., In the special case in which c  0, the Taylor series becomes, ⬁, , f(x)  a, n0, , f (n) (0) n, f ⬙(0) 2 f ‡(0) 3 p, x  f(0)  f ¿(0)x , x , x , n!, 2!, 3!, , (2)

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9.8, , Taylor and Maclaurin Series, , 795, , This series is just the Taylor series of f centered at the origin. It is called the Maclaurin series of f in honor of the Scottish mathematician Colin Maclaurin (1698–1746)., Note Theorem 1 states that if a function f has a power series representation at c, then, the (unique) series must be the Taylor series at c. The converse is not necessarily true., Given a function f with derivatives of all orders at c, we can compute the Taylor coefficients of f at c,, f (n) (c), n!, , n  0, 1, 2, p, , and, therefore, the Taylor series of f at c (Equation (1)). But the series that is obtained, formally in this fashion need not represent f. Situations such as these, however, are rare., (We give an example of such a function in Exercise 75.) In view of this we will assume,, in the rest of this section, that the Taylor series of a function does represent the function, unless otherwise noted., , EXAMPLE 1 Let f(x)  ex. Find the Maclaurin series of f, and determine its radius, of convergence., Solution The derivatives of f(x)  ex are f ¿(x)  ex, f ⬙(x)  ex, and, in general,, f (n) (x)  ex, where n  1. So, f(0)  1,, , f ¿(0)  1,, , f ⬙(0)  1,, , f (n) (0)  1,, , p,, , p, , Therefore, if we use Equation (2), the Maclaurin series of f (the Taylor series of f, at 0) is, ⬁, , ⬁, f (n) (0) n, 1 n, x2, x3 p xn p, x, , x, , 1, , x, , ,  , , a n!, a, 2!, 3!, n!, n0, n0 n!, , To determine the radius of convergence of the power series, we use the ratio test with, u n  x n>n!. Since, lim `, , n→⬁, , 冟x冟, u n1, x n1, n!, `  lim `, ⴢ n `  lim, 0, un, n→⬁ (n  1)! x, n→⬁ n  1, , we conclude that the radius of convergence of the series is R  ⬁ ., , EXAMPLE 2 Find the Taylor series for f(x)  ln x at 1, and determine its interval of, convergence., Solution, , We compute the values of f and its derivatives at 1. Thus,, f(x)  ln x, f ¿(x) , , 1,  x 1, x, , f ⬙(x)  x 2, f ‡(x)  2x, , 3, , f (4) (x)  3 ⴢ 2x 4, o, f (n) (x)  (1)n1(n  1)! x n, , f(1)  ln 1  0, f ¿(1)  1, f ⬙(1)  1, f ‡(1)  2, f (4) (1)  3 ⴢ 2, o, f (n) (1)  (1)n1(n  1)!

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796, , Chapter 9 Infinite Sequences and Series, , Then using Equation (1), we obtain the Taylor series of f(x)  ln x:, ⬁, , a, n0, , f (n) (1), f ⬙(1), f ‡(1), (x  1)n  f(1)  f ¿(1)(x  1) , (x  1)2 , (x  1)3  p, n!, 2!, 3!,  (x  1) , , 1, 2, 3!, (x  1)2 , (x  1)3 , (x  1)4  p, 2!, 3!, 4!, ,  (x  1) , , (x  1)2, (x  1)3, (x  1)4 p, , , , 2, 3, 4, , ⬁, , (x  1)n, n, ,  a (1)n1, n1, , To find the interval of convergence of the series, we use the Ratio Test with, u n  (1)n1(x  1)n>n. Since, lim `, , n→⬁, , (1)n (x  1)n1, u n1, n, `  lim `, ⴢ, `, n1, un, n→⬁, n1, (1) (x  1)n,  lim 冟 x  1 冟 a, n→⬁, , n, b  冟 x  1 冟 lim, n1, n→⬁, , 1, 1, 1, n, ,  冟x  1冟, , we see that the series converges for x in the interval (0, 2). Next, we notice that if, x  0, the series becomes, ⬁, , ⬁, (1)2n1, 1, , , a, a, n, n1, n1 n, , Since this is the negative of the harmonic series, it is divergent. If x  2, the series, becomes, ⬁, , (1)n1, a, n, n1, This is the alternating harmonic series and, hence, is convergent. Therefore, the Taylor series for f(x)  ln x at 1 has interval of convergence (0, 2]., , EXAMPLE 3 Find the Maclaurin series of f(x)  sin x, and determine its interval of, convergence., Solution To find the Maclaurin series of f(x)  sin x, we compute the values of f and, its derivatives at x  0. We obtain, f(x)  sin x, , f(0)  0, , f ¿(x)  cos x, , f ¿(0)  1, , f ⬙(x)  sin x, , f ⬙(0)  0, , f ‡(x)  cos x, , f ‡(0)  1, , f (4) (x)  sin x, , f (4) (0)  0

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9.8, , Taylor and Maclaurin Series, , 797, , We need not go further, since it is clear that successive derivatives of f follow this same, pattern. Then, using Equation (2), we obtain the Maclaurin series of f(x)  sin x:, ⬁, , f (n) (0) n, f ⬙(0) 2 f ‡(0) 3 f (4) (0) 4 p, x, , f(0), , f, ¿(0)x, , x , x , x , a n!, 2!, 3!, 4!, n0, x, , x3, x5, x7 p, , , , 3!, 5!, 7!, , ⬁, (1)n,  a, x 2n1, (2n, , 1)!, n0, , To find the interval of convergence of the series, we use the Ratio Test with, u n  (1)nx 2n1>(2n  1)!. Since, lim `, , n→⬁, , (1)n1x 2n3 (2n  1)!, u n1, `  lim `, ⴢ, `, un, n→⬁, (2n  3)!, (1)nx 2n1,  lim, , n→⬁, , 冟 x 冟2, (2n  2)(2n  3), , 01, , we conclude that the interval of convergence of the series is (⬁, ⬁)., , EXAMPLE 4 Find the Maclaurin series of f(x)  cos x., Solution We could proceed as in Example 3, but it is easier to make use of Theorem 2 of Section 9.7 to differentiate the expression for sin x that we obtained in Example 3. Thus,, f(x)  cos x , 1, , d, x3, x5, x7 p, d, (sin x) , ax , , ,  b, dx, dx, 3!, 5!, 7!, , x2, x4, x6 p, , , , 2!, 4!, 6!, , ⬁, (1)n 2n,  a, x, n0 (2n)!, , Since the Maclaurin series for sin x converges for all x, Theorem 2 of Section 9.7 tells, us that this series converges in (⬁, ⬁) as well., , EXAMPLE 5 Find the Maclaurin series for f(x)  (1  x)k, where k is a real number., Solution, , We compute the values of f and its derivatives at x  0, obtaining, , f(x)  (1  x)k, , f(0)  1, , f ¿(x)  k(1  x)k1, f ⬙(x)  k(k  1)(1  x), , f ¿(0)  k, k2, , f ‡(x)  k(k  1)(k  2)(1  x)k3, o, f (n) (x)  k(k  1) p (k  n  1)(1  x)kn, , f ⬙(0)  k(k  1), f ‡(0)  k(k  1)(k  2), o, f (n) (0)  k(k  1) p (k  n  1)

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798, , Chapter 9 Infinite Sequences and Series, , So the Maclaurin series of f(x)  (1  x)k is, ⬁, , f (n) (0) n, f ⬙(0) 2 f ‡(0) 3 p, a n! x  f(0)  f ¿(0)x  2! x  3! x , n0, k(k  1) 2 k(k  1)(k  2) 3 p, x , x , 2!, 3!, ,  1  kx , ⬁, ,  a, n0, , k(k  1)(k  2) p (k  n  1) n, x, n!, , Observe that if k is a positive integer, then the series is infinite (by the Binomial Theorem), and so it converges for all x., If k is not a positive integer, then we use the Ratio Test to find the interval of convergence. Denoting the nth term of the series by u n, we find, lim `, , n→⬁, , k(k  1) p (k  n  1)(k  n)x n1, u n1, n!, `  lim `, ⴢ, `, p, un, n→⬁, (n  1)!, k(k  1) (k  n  1)x n, k,  1`, 冟k  n冟, n, 冟x冟  冟x冟, 冟, 冟,  lim, x  lim, n→⬁ n  1, n→⬁, 1, 1, n, `, , and we see that the series converges for x in the interval (1, 1)., The series in Example 5 is called the binomial series., , The Binomial Series, If k is any real number and 冟 x 冟  1, then, (1  x)k  1  kx , , ⬁, k(k  1) 2 k(k  1)(k  2) 3 p, k, x , x   a a bx n (3), 2!, 3!, n0 n, , Notes, 1. The coefficients in the binomial series are referred to as binomial coefficients, and are denoted by, k(k  1) p (k  n  1), k, a b, n, n!, , n  1,, , k, a b1, 0, , 2. If k is a positive integer and n  k, then the binomial coefficient contains a factor, (k  k), so 1 kn 2  0 for n  k. The binomial series then reduces to a polynomial, of degree k:, (1  x)k  1  kx , , k, k(k  1) 2 p, k, x   x k  a a bx n, n, 2!, n0, , In other words, the expression (1  x)k can be represented by a finite sum if k is, a positive integer and by an infinite series if k is not a positive integer. Thus, we, can view the binomial series as an extension of the Binomial Theorem to the case, in which k is not a positive integer.

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9.8, , Taylor and Maclaurin Series, , 799, , 3. Even though the binomial series always converges for ⫺1 ⬍ x ⬍ 1, its convergence at the endpoints x ⫽ ⫺1 or x ⫽ 1 depends on the value of k. It can be, shown that the series converges at x ⫽ 1 if ⫺1 ⬍ k ⬍ 0 and at both endpoints, x ⫽ ⫾1 if k ⱖ 0., 4. We have derived Equation (3) under the assumption that (1 ⫹ x)k has a power, series representation. In Exercise 78 we outline a procedure for deriving, Equation (3) without this assumption., , EXAMPLE 6 Find a power series representation for the function f(x) ⫽ 11 ⫹ x., Solution, , Using Equation (3) with k ⫽ 12, we obtain, , f(x) ⫽ (1 ⫹ x), ⫹, y, , P2, P3, , 1, , P1, , ⫽1⫹, , ⫽1⫹, 0, , 1, ⫽1⫹ x⫹, 2, , ⫺ 12 p, , 1 x, , 1, , 1 1, 2 2, , 1 12 ⫺ n ⫹ 1 2, , n!, , ⫺ 12, , 2!, , x ⫹, 2, , 1, , 1 1, 2 2, , ⫺ 1 21 12 ⫺ 2 2, 3!, , x3 ⫹ p, , xn ⫹ p, , 1, 1, 1ⴢ3 3 p, x⫺, x2 ⫹, x ⫹, 2, 2, 2ⴢ2, 3! 23, , ⫹ (⫺1)n⫹1, , f(x) ⫽ √1 ⫹ x, ⫺1, , 1, , 1 1, 2 2, , 1>2, , 1 ⴢ 3 ⴢ 5 ⴢ p ⴢ (2n ⫺ 3) n p, x ⫹, n! 2n, , ⬁, 1 ⴢ 3 ⴢ 5 ⴢ p ⴢ (2n ⫺ 3) n, 1, x ⫹ a (⫺1)n⫹1, x, 2, n! 2n, n⫽2, , This representation is valid for 冟 x 冟 ⱕ 1., The graph of f and the first three partial sums P1(x) ⫽ 1, P2(x) ⫽ 1 ⫹ 12 x, and, P3(x) ⫽ 1 ⫹ 12 x ⫺ 18 x 2 are shown in Figure 1. Observe that the partial sums of f,, Pn(x), approximate f better and better in the interval of convergence of the series as n, increases., , FIGURE 1, The graphs of f(x) ⫽ 11 ⫹ x and the, first three partial sums of the binomial, series, , Techniques for Finding Taylor Series, The Taylor series of a function can always be found by using Equation (1). But as, Examples 7, 8, and 9 of Section 9.7 and Example 4 of this section show, it is often, easier to find the series by algebraic manipulation, differentiation, or integration of, some well-known series. We now elaborate further on such techniques. First, we list, some common functions and their power series representations in Table 1., TABLE 1, Maclaurin Series, , Interval of Convergence, ⬁, , 1., , 1, ⫽ 1 ⫹ x ⫹ x2 ⫹ x3 ⫹ p ⫽ a xn, 1⫺x, n⫽0, , 2. ex ⫽ 1 ⫹ x ⫹, 3. sin x ⫽ x ⫺, 4. cos x ⫽ 1 ⫺, , ⬁, x2, xn, x3 p, ⫹, ⫹ ⫽ a, 2!, 3!, n⫽0 n!, , ⬁, x3, x5, x7 p, x 2n⫹1, ⫹, ⫺, ⫹ ⫽ a (⫺1)n, 3!, 5!, 7!, (2n ⫹ 1)!, n⫽0, ⬁, x4, x6 p, x 2n, x2, ⫹, ⫺, ⫹ ⫽ a (⫺1)n, 2!, 4!, 6!, (2n)!, n⫽0, , (⫺1, 1), (⫺⬁, ⬁), (⫺⬁, ⬁), (⫺⬁, ⬁), (continued)

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800, , Chapter 9 Infinite Sequences and Series, , TABLE 1 (continued), Maclaurin Series, 5. ln(1  x)  x , , Interval of Convergence, , ⬁, x2, x3, x4 p, xn, , ,   a (1)n1, n, 2, 3, 4, n1, , (1, 1], , 6. sin1 x  x , , ⬁, (2n)! x, x3, 1 ⴢ 3x 5, , p a n 2, 2ⴢ3, 2ⴢ4ⴢ5, (2, n!) (2n  1), n0, , [1, 1], , 7. tan1 x  x , , ⬁, x3, x5, x7 p, x 2n1, , ,   a (1)n, 3, 5, 7, 2n  1, n0, , [1, 1], , 2n1, , ⬁, k(k  1) 2 k(k  1)(k  2) 3, k, x , x p, 8. (1  x)k  a a bx n  1  kx , 2!, 3!, n0 n, , (1, 1), , All of the formulas in the table except Formulas (5) and (6) have been derived in, this and the previous sections. (See note on page 795.) Formula (5) follows from the, result of Example 8 in Section 9.7 by replacing x by x. Formula (6) will be derived, in Example 14., , EXAMPLE 7 Find the Taylor series representation of f(x) , Solution, , 1, at x  2., 1x, , We first rewrite f(x) so that it includes the expression (x  2). Thus,, , f(x) , , 1, 1, , , 1x, 3  (x  2), , 1, 3c1  a, , x2, bd, 3, , , , 1, ⴢ, 3, , 1, 1a, , x2, b, 3, , Then, using Formula (1) in Table 1 with x replaced by (x  2)>3, we obtain, f(x) , , 1, 1, •, x2 ¶, 3 1  ca, bd, 3, , , , 1, x2, x2 2, x2 3 p, e1  ca, b d  ca, b d  ca, bd  f, 3, 3, 3, 3, , , , 1, x2, x2 2, x2 3 p, c1  a, ba, b a, b  d, 3, 3, 3, 3, , , , ⬁, (x  2)n, 1, 1, 1, 1,  2 (x  2)  3 (x  2)2  4 (x  2)3  p  a (1)n, 3, 3, 3, 3, 3n1, n0, , The series converges for 冟 (x  2)>3 冟  1, that is, 冟 x  2 冟  3 or 1  x  5. You can, verify that the series diverges at both endpoints., , EXAMPLE 8 Find the Maclaurin series for f(x)  x 2 sin 2x., Solution, , If we replace x by 2x in Formula (3) in Table 1, we obtain, sin 2x  (2x) ,  2x , , (2x)3, (2x)5, (2x)7 p, , , , 3!, 5!, 7!, , ⬁, 23x 3, 25x 5, 27x 7 p, 22n1x 2n1, , ,   a (1)n, 3!, 5!, 7!, (2n  1)!, n0

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9.8, , Taylor and Maclaurin Series, , 801, , which is valid for all x in (⬁, ⬁). Therefore, using Theorem 4a of Section 9.2, we, obtain, f(x)  x 2 sin 2x  x 2 a2x ,  2x 3 , , 23x 3, 25x 5, 27x 7 p, , ,  b, 3!, 5!, 7!, , 23x 5, 25x 7, 27x 9 p, , , , 3!, 5!, 7!, , ⬁, 22n1x 2n3,  a (1)n, (2n  1)!, n0, , which converges for all x in (⬁, ⬁)., The next example shows how the use of trigonometric identities can help us find, the Taylor series of a trigonometric function., , EXAMPLE 9 Find the Taylor series for f(x)  sin x at x  p>6., Solution, , We write, f(x)  sin x  sinc ax ,  sinax , , , p, p, b d, 6, 6, , p, p, p, p, bcos  cosax  bsin, 6, 6, 6, 6, , 13, p, 1, p, sinax  b  cosax  b, 2, 6, 2, 6, , Then using Formulas 3 and 4 with x  (p>6) in place of x, we obtain, f(x) , , (1)n, 13 ⬁, p 2n1 1 ⬁ (1)n, p 2n, ax, , b,  a, ax  b, a, 2 n0 (2n  1)!, 6, 2 n0 (2n)!, 6, , which converges for all x in (⬁, ⬁)., The power series representations of certain functions can also be found by adding,, multiplying, or dividing the Maclaurin or Taylor series of some familiar functions as, the following examples show., , EXAMPLE 10 Find the Maclaurin series representation for f(x)  sinh x., Solution, , We have, , f(x)  sinh x , , , 1 x, 1, 1, (e  ex)  ex  ex, 2, 2, 2, 1, x2, x3 p, 1, x2, x3 p, a1  x , ,  b  a1  x , ,  b, 2, 2!, 3!, 2, 2!, 3!, , x, , ⬁, x3, x5 p, x 2n1, ,   a, 3!, 5!, n0 (2n  1)!, , Since the Maclaurin series of both ex and ex converge for x in (⬁, ⬁), we see that, this representation of sinh x is also valid for all values of x.

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802, , Chapter 9 Infinite Sequences and Series, , EXAMPLE 11 Find the first three terms of the Maclaurin series representation for, f(x)  ex cos x., Solution, , Using Formulas (2) and (4) in Table 1, we can write, f(x)  ex cos x  a1  x , , x2, x3 p, x2, x4, ,  b a1 , ,  pb, 2, 6, 2, 24, , Multiplying and collecting like terms, we obtain, f(x)  (1)a1 , , , x2, x2, x4, x3, x2 p, a1 , ,  pb , a1 ,  bp, 2, 2, 24, 6, 2, , 1, , , x2, x4, x2, x4, ,  p b  xa1 , ,  pb, 2, 24, 2, 24, , x2, x4, x3, x5, x2, x4, , px, , p, , 2, 24, 2, 24, 2, 4, , x6, x3, x5, p, , p, 48, 6, 12, , 1x, , x3 p, , 3, , EXAMPLE 12 Find the first three terms of the Maclaurin series representation for, f(x)  tan x., Solution, , Using Formulas (3) and (4) in Table 1, we have, , f(x)  tan x , , sin x, , cos x, , x, , x3, x5 p, , , 3!, 5!, , 1, , x2, x4 p, , , 2!, 4!, , By long division we find, x  13 x 3 , 1  12 x 2  241 x 4  p, , b, , 2 5, 15 x, , p, , 1, x  16 x 3  120, x5  p, , x  12 x 3 , , 1 5, 24 x, , p, , 1 3, 3x, , , , 1 5, 30 x, , p, , 1 3, 3x, , , , 1 5, 6x, , p, , 2 5, 15 x, , p, , Therefore,, f(x)  tan x  x , , 1 3, 2 5 p, x , x , 3, 15, , In both Examples 11 and 12 we computed only the first three terms of each series. In, practice, the retention of just the first few terms of a series is sufficient to obtain an, acceptable approximation to the solution of a problem.