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Sequence and Series, Engineering Mathematics - I, , 2, , 1.1 Sequence, A function f:N → S, where S is any nonempty set is called a Sequence, i.e., for each n ∈ N, ∃ a unique element f(n) ∈ S. The sequence is written as f(1), f(2),, f(3), ......f(n)...., and is denoted by {f(n)}, or <f(n)>, or (f(n)). If f(n) = an , the sequence is, written as a1 , a2 .....an and denoted by , {an } or < an > or ( an ) . Here f(n) or an are the, , nth terms of the Sequence., Ex. 1., , 1 , 4 , 9 , 16 ,......... n 2 ,.....(or) < n 2 >, N, , S, , 2, 3, , 1, 4, 9, , n, , n2, , ., ., , ., ., , Ex. 2., , 1 1 1, 1, ⎛ 1 ⎞, , 3 , 3 ,..... 3 ....(or ) ⎜ 3 ⎟, 3, 1 2 3, n, ⎝n ⎠, , Ex. 3., , 1, 1, 1......1..... or <1>, , ., ., ., .

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Sequences and Series, , Ex 4:, , 3, , ( −1), , 1 , –1, 1, –1, ......... or, , n −1, , Note : 1. If S ⊆ R then the sequence is called a real sequence., 2. The range of a sequence is almost a countable set., 1.1.1 Kinds of Sequences, 1. Finite Sequence: A sequence < an > in which an = 0 ∀n > m ∈ N is said to, , be a finite Sequence. i.e., A finite Sequence has a finite number of terms., 2. Infinite Sequence: A sequence, which is not finite, is an infinite sequence., 1.1.2 Bounds of a Sequence and Bounded Sequence, 1. If ∃ a number ‘M’ ∋ an ≤ M, ∀n ∈ N, the Sequence < an > is said to be, , bounded above or bounded on the right., 1 1, Ex. 1, , , ,....... here an ≤ 1 ∀n ∈ N, 2 3, 2. If ∃ a number ‘m’ ∋ an ≥ m, ∀n ∈ N, the sequence < an > is said to be, bounded below or bounded on the left., Ex. 1 , 2 , 3 ,.....here an ≥ 1 ∀n ∈ N, 3. A sequence which is bounded above and below is said to be bounded., 1⎞, n⎛, Ex. Let an = ( −1) ⎜ 1 + ⎟, ⎝ n⎠, , n, , 1, , 2, , 3, , 4, , ......, , an, , -2, , 3/2, , -4/3, , 5/4, , ......

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Engineering Mathematics - I, , 4, , From the above figure (see also table) it can be seen that m = –2 and M =, , ∴ The sequence is bounded., , 3, ., 2, , 1.1.3 Limits of a Sequence, , A Sequence < an > is said to tend to limit ‘l’ when, given any + ve number ' ∈ ',, however small, we can always find an integer ‘m’ such that an − l <∈, ∀n ≥ m , and we, write Lt an = l or an → l, n →∞, , Ex., , If an =, , 1, n2 + 1, then < an >→ ., 2, 2, 2n + 3, , 1.1.4 Convergent, Divergent and Oscillatory Sequences, 1. Convergent Sequence: A sequence which tends to a finite limit, say ‘l’ is, called a Convergent Sequence. We say that the sequence converges to ‘l’, 2. Divergent Sequence: A sequence which tends to ±∞ is said to be Divergent, (or is said to diverge)., 3. Oscillatory Sequence: A sequence which neither converges nor diverges ,is, called an Oscillatory Sequence., 3 4 5, 1, Ex. 1. Consider the sequence 2 , , , ,..... here an = 1 +, n, 2 3 4, The sequence < an > is convergent and has the limit 1, 1, 1, 1, 1, an − 1 = 1 + − 1 = and <∈ whenever n >, n, n, n, ∈, 1, Suppose we choose ∈= .001 , we have < .001 when n > 1000., n, 1, n, < an > converges to 3., Ex. 2. If an = 3 + ( −1), 'n'

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Sequences and Series, , 5, , Ex. 3., , If an = n 2 + ( −1) .n, < an > diverges., , Ex. 4., , If an =, , n, , 1, n, + 2 ( −1) , < an > oscillates between -2 and 2., n, , 1.2 Infinite Series, , If < un > is a sequence, then the expression u1 + u2 + u3 + ........ + un + ..... is called an, ∞, , infinite series. It is denoted by Σ un or simply Σun, n =1, , The sum of the first n terms of the series is denoted by sn, i.e.,, , sn = u1 + u2 + u3 + ...... + un ; s1 , s2 , s3 ,....sn are called partial sums., , 1.2.1 Convergent, Divergent and Oscillatory Series, , Let Σun be an infinite series. As n → ∞, there are three possibilities., (a) Convergent series: As n → ∞, sn → a finite limit, say ‘s’ in which case the, series is said to be convergent and ‘s’ is called its sum to infinity., , Thus Lt sn = s (or) simply Ltsn = s, n →∞, , ∞, , This is also written as u1 + u2 + u3 + ..... + un + ...to∞ = s. (or) Σ un = s (or), n =1, , simply Σun = s., (b) Divergent series: If sn → ∞ or −∞ , the series said to be divergent., (c) Oscillatory Series: If sn does not tend to a unique limit either finite or infinite it, is said to be an Oscillatory Series., Note: Divergent or Oscillatory series are sometimes called non convergent series., 1.2.2 Geometric Series, , The series, 1 + x + x 2 + .....x n −1 + ... is, (i) Convergent when x < 1 , and its sum is, (ii) Divergent when x ≥ 1 ., , 1, 1− x, , (iii) Oscillates finitely when x = -1 and oscillates infinitely when x < -1., Proof: The given series is a geometric series with common ratio ‘x’, 1 − xn, ∴ sn =, when x ≠ 1, [By actual division – verify], 1− x

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Engineering Mathematics - I, , 6, , (i) When x < 1:, ⎛ xn ⎞, 1, ⎛ 1 ⎞, ⎡⎣since x n → 0 as n → ∞ ⎤⎦, −, =, Lt sn = Lt ⎜, Lt, ⎜, ⎟, ⎟, n →∞, n →∞ 1 − x, ⎝, ⎠ n →∞ ⎝ 1 − x ⎠ 1 − x, 1, ∴ The series converges to, 1− x, n, x −1, and sn → ∞ as n → ∞, (ii) When x ≥ 1: sn =, x −1, ∴ The series is divergent., (iii) When x = –1: when n is even, sn → 0 and when n is odd, sn → 1, ∴ The series oscillates finitely., (iv) When x < −1, sn → ∞ or −∞ according as n is odd or even., ∴ The series oscillates infinitely., , 1.2.3 Some Elementary Properties of Infinite Series, 1. The convergence or divergence of an infinites series is unaltered by an addition or, deletion of a finite number of terms from it., 2. If some or all the terms of a convergent series of positive terms change their signs,, the series will still be convergent., 3. Let Σun converge to ‘s’, , Let ‘k’ be a non – zero fixed number. Then Σkun converges to ks., Also, if Σun diverges or oscillates, so does Σkun, 4. Let Σun converge to ‘l’ and Σvn converge to ‘m’. Then, (i) Σ(un + vn ) converges to ( l + m ) and (ii) Σ(un + vn ) converges to ( l – m ), 1.2.4 Series of Positive Terms, , Consider the series in which all terms beginning from a particular term are +ve., Let the first term from which all terms are +ve be u1 ., Let Σun be such a convergent series of +ve terms. Then, we observe that the, convergence is unaltered by any rearrangement of the terms of the series., 1.2.5 Theorem, , If Σun is convergent, then Lt un = 0 ., n →∞, , Proof :, , sn = u1 + u2 + ...... + un, sn −1 = u1 + u2 + ...... + un −1, , so that, un = sn − sn −1

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Sequences and Series, , 7, , Suppose Σun = l then Lt sn = l and Lt sn −1 = l, n →∞, , ∴ Lt un = Lt ( sn − sn −1 ) ;, n →∞, n →∞, , n →∞, , Lt sn − Lt sn −1 = l − l = 0, , n →∞, , n →∞, , Note: The converse of the above theorem need not be always true. This can be, observed from the following examples., 1 1, 1, 1, (i) Consider the series, 1 + + + ....... + + .... ; un = , Lt un = 0, n n →∞, 2 3, n, , But from p-series test (1.3.1) it is clear that Σ, (ii), , Consider the series,, , 1, is divergent., n, , 1 1 1, 1, + 2 + 2 + ..... + 2 + ......, 2, 1 2 3, n, , 1, 1, , Lt u = 0, by p series test, clearly Σ 2 converges,, 2 n →∞ n, n, n, Note : If Lt un ≠ 0 the series is divergent;, un =, , n →∞, , Ex., , un =, , 2n − 1, , here Lt un = 1 ∴, n →∞, 2n, , Σun is divergent., , 1.3 Tests for the Convergence of an Infinite Series, In order to study the nature of any given infinite series of +ve terms regarding, convergence or otherwise, a few tests are given below., 1.3.1 P-Series Test, ∞, , 1, 1, 1, 1, = p + p + p + ......, is, p, n, 1, 2, 3, (i) Convergent when p > 1, and (ii) Divergent when p ≤ 1 ., , The infinite series, Σ, , n =1, , Proof :, Case (i) Let, , p > 1,3 p > 2 p ; ⇒, , p > 1;, , ∴, Similarly,, , 1, 1, < p, p, 3, 2, , 1, 1, 1, 1, 2, + p < p+ p = p, p, 2, 3, 2, 2, 2, 1, 1, 1, 1, 1, 1, 1, 1, 4, + p + p+ p < p+ p + p+ p = p, p, 4, 5, 6, 7, 4, 4, 4, 4, 4, , 1, 1, 1, 8, + p + .... + p < p , and so on., p, 8, 9, 16, 8, Adding we get, , (JNTU 2002, 2003)

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Engineering Mathematics - I, , 8, , Σ, , 1, 2, 4, 8, < 1 + p + p + p + ...., p, n, 2, 4, 8, , 1, 1, 1, 1, < 1 + ( p −1) + 2( p −1) + 3( p −1) + ......, p, n, 2, 2, 2, The RHS of the above inequality is an infinite geometric series with common, 1, ratio p −1 < 1( since p > 1) The sum of this geometric series is finite., 2, Σ, , i.e.,, , ∞, , 1, is also finite., n =1 n p, , Hence Σ, , ∴ The given series is convergent., 1, 1 1 1, = 1 + + + + ......, np, 2 3 4, 1 1 1 1 1, + > + =, We have,, 3 4 4 4 2, 1 1 1 1 1 1 1 1 1, + + + > + + + =, 5 6 7 8 8 8 8 8 2, 1 1, 1, 1 1, 1 1, + + ....... > + + ..... = and so on, 9 10, 16 16 16, 16 2, 1, ⎛1 1⎞ ⎛1 1 1 1⎞, Σ p = 1 + ⎜ + ⎟ + ⎜ + + + ⎟ + ....., ∴, n, ⎝ 2 3⎠ ⎝ 4 5 6 7 ⎠, 1 1 1, ≥ 1 + + + + ....., 2 2 2, The sum of RHS series is ∞, n −1 n +1, ⎛, ⎞, and Lt sn = ∞ ⎟, =, ⎜ since sn = 1 +, n →∞, 2, 2, ⎝, ⎠, , Case (ii) Let p =1;, , Σ, , ∞, , 1, ( p = 1 ) diverges., n =1 n p, , ∴ The sum of the given series is also ∞ ; ∴ Σ, , 1, 1, 1, = 1 + p + p + ....., p, n, 2, 3, 1, 1 1 1, p < 1, p > , p > , ...... and so on, Since, 2, 2 3, 3, 1, 1 1 1, ∴, Σ p > 1 + + + + ......., n, 2 3 4, From the Case (ii), it follows that the series on the RHS of above inequality is, divergent., , Case (iii) Let p<1,, , Σ

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Sequences and Series, , ∴, , 9, , Σ, , 1, is divergent , when P < 1, np, , Note: This theorem is often helpful in discussing the nature of a given infinite series., 1.3.2 Comparison Tests, 1. Let Σun and Σvn be two series of +ve terms and let Σvn be convergent., Then Σun converges,, (a) If un ≤ vn , ∀ n ∈ N, u, (b) or n ≤ k ∀n ∈ N where k is > 0 and finite., vn, u, (c) or n → a finite limit > 0, vn, Proof : (a) Let Σvn = l (finite), Then, u1 + u2 + ..... + un + ...... ≤ v1 + v2 + .....vn + ..... ≤ l > 0, , Since l is finite it follows that Σun is convergent, un, ≤ k ⇒ un ≤ kvn , ∀n ∈ N , since Σvn is convergent and k (>0) is finite,, vn, Σkvn is convergent ∴ Σun is convergent., u, u, (d) Since Lt n is finite, we can find a +ve constant k ,∋ n < k ∀n ∈ N, n →∞ v, vn, n, ∴ from (2) , it follows that Σun is convergent, 2. Let Σun and Σvn be two series of +ve terms and let Σvn be divergent. Then, Σun diverges,, * 1. If un ≥ vn , ∀n ∈ N, u, or * 2. If n ≥ k , ∀n ∈ N where k is finite and ≠ 0, vn, u, or * 3. If Lt n is finite and non-zero., n →∞ v, n, Proof:, 1. Let M be a +ve integer however large it may be. Science Σvn is divergent, a, number m can be found such that, v1 + v2 + ..... + vn > M , ∀n > m, , (c), , ∴, ∴, , u1 + u2 + ..... + un > M , ∀n > m ( un ≥ vn ), Σun is divergent

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Engineering Mathematics - I, , 10, , 2. u1 ≥ kvn ∀n, Σvn is divergent ⇒ Σkvn is divergent, ∴, Σun is divergent, u, u, 3. Since Lt n is finite, a + ve constant k can be found such that n > k , ∀n, n →∞ v, vn, n, (probably except for a finite number of terms ), ∴ From (2), it follows that Σun is divergent., Note :, (a), , In (1) and (2), it is sufficient that the conditions with * hold ∀n > m ∈ N, Alternate form of comparison tests : The above two types of comparison tests, 2.8.(1) and 2.8.(2) can be clubbed together and stated as follows :, , un, = k, where k is, n →∞ v, n, , If Σun and Σvn are two series of + ve terms such that Lt, , non- zero and finite, then Σun and Σvn both converge or both diverge., (b), , 1. The above form of comparison tests is mostly used in solving problems., 2. In order to apply the test in problems, we require a certain series Σvn whose, nature is already known i.e., we must know whether Σvn is convergent are, divergent. For this reason, we call Σvn as an ‘auxiliary series’., 3. In problems, the geometric series (1.2.2.) and the p-series (1.3.1) can be, conveniently used as ‘auxiliary series’., , Solved Examples, EXAMPLE 1, Test the convergence of the following series:, 3 4 5, 6, 4 5 6 7, +, + ..... (b), + + + + ....., (a) + +, 1 8 27 64, 1 4 9 16, , ∞, , (c) Σ ⎡( n 4 + 1), n =1 ⎢, ⎣, , 14, , − n⎤, ⎥⎦, , SOLUTION, (a), , Step 1: To find "un " the nth term of the given series. The numerators 3, 4, 5,, 6......of the terms, are in AP., nth term tn = 3 + ( n − 1) .1 = n + 2, , Denominators are 13 , 23 ,33 , 43.....nth term = n3 ; ∴ un =, , n+2, n3, , Step 2: To choose the auxiliary series Σvn . In un , the highest degree of n in the, numerator is 1 and that of denominator is 3.

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Sequences and Series, , ∴ we take, vn =, Step 3: Lt, , n →∞, , 11, , 1, 1, = 2, 3−1, n, n, , un, n+2 2, n+2, ⎛ 2⎞, = Lt, × n = Lt, = Lt ⎜1 + ⎟ = 1, which is non- zero and, 3, n, n, n, →∞, →∞, →∞, vn, n, n, ⎝ n⎠, , finite., un, =1, n →∞ v, n, , Step 4: Conclusion: Lt, , ∴ Σun and Σvn both converge or diverge (by comparison test). But Σvn = Σ, convergent by p-series test (p = 2 > 1); ∴ Σun is convergent., (b), , 1, is, n2, , 4 5 6 7, + + + + ....., 1 4 9 16, , Step 1: 4 , 5, 6, 7, .....in AP , tn = 4 + ( n − 1)1 = n + 3, , ∴ un =, , n+3, n2, , 1, be the auxiliary series, n, u, ⎛ n +3⎞, ⎛ 3⎞, Step 3: Lt n = Lt ⎜ 2 ⎟ × n = Lt ⎜1 + ⎟ = 1 , which is non-zero and finite., →∞, n →∞ v, n →∞, n, ⎝ n ⎠, ⎝ n⎠, n, Step 4: ∴ By comparison test, both Σun and Σvn converge are diverge together., 1, But Σvn = Σ is divergent, by p-series test (p = 1); ∴ Σun is divergent., n, Step 2: Let Σvn =, , ∞, , (c), , Σ ⎡( n + 1), ⎣⎢, , n =1, , 4, , 14, , 1, 1, ⎡, ⎤, 4, 4, ⎧, ⎫, 1, 1, ⎛, ⎞, ⎛, ⎞, 4, ⎢, ⎤, − n = ⎨n ⎜1 + 4 ⎟⎬ – n = n ⎜1 + 4 ⎟ – 1⎥, ⎢⎝ n ⎠, ⎥, ⎦⎥ ⎩ ⎝ n ⎠⎭, ⎣⎢, ⎦⎥, , ⎡, ⎤, 1⎛1 ⎞, − 1⎟, ⎜, ⎢, ⎥, 1, 3, 4 4 ⎠ 1, ⎡ 1, ⎤, . 8 + ..... − 1⎥ = n ⎢ 4 −, = n ⎢1 + 4 + ⎝, + .....⎥, 8, 4, 2!, 4, 32, n, n, n, n, ⎢, ⎥, ⎣, ⎦, ⎢⎣, ⎥⎦, 1, 3, 1 ⎡1, 3, ⎤, = 3−, + .... = 3 ⎢ −, + .....⎥, n ⎣ 4 32n 4, 4n 32n 7, ⎦, 1, Here it will be convenient if we take vn = 3, n

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Engineering Mathematics - I, , 12, , un, 1, ⎛1, ⎞ 1, = Lt ⎜ −, + ..... ⎟ = , which is non-zero and finite, 4, n →∞ v, n →∞ 4, 32n, ⎝, ⎠ 4, n, Lt, , ∴ By comparison test, Σun and Σvn both converge or both diverge. But by pseries test Σvn =, , 1, is convergent. (p = 3 > 1); ∴ Σun is convergent., n3, , EXAMPLE 2, , If u n =, , 3, 4, , 3n 2 + 1, , 2n3 + 3n + 5, , show that Σun is divergent, , SOLUTION, , As n increases, un approximates to, 3, 4, , 3n 2, 2n3, , =, , 3, 2, , 1, , 3, , 1, , ×, , 4, , ∴ If we take vn =, , n, , 2, , n, , 3, , 3, , =, , 4, , [(or) Hint: Take vn =, , 1, 12, , 1, n, , 2, , 1, , 3, , 1, , ., 4, , 1, n, , 1, 12, , 1, , 1, n, , 3, , l1 − l2, , u, 33, , Lt n = 1 which is finite., n →∞ v, n, 2 4, , where l1 and l2 are indices of ‘n’ of the largest terms, 1, , in denominator and nominator respectively of un . Here vn =, n, , 3 2, −, 4 3, , =, , 1, n, , 1, 12, , ], , By comparison test, Σvn and Σun converge or diverge together. But Σvn = Σ, divergent by p – series test ( since p =, , ∴ Σun is divergent., , 1, n, , 1, <1), 12, , EXAMPLE 3, , 1, 2, 3, 4, +, +, +, + ......, 2, 3, 4, 5, , Test for the convergence of the series., SOLUTION, , Here, un =, , n, ;, n +1, , 1, , Take vn =, , n, , 1 1, −, 2 2, , =, , u, 1, 1, = 1 (finite), = 1 , Lt n = Lt, 0, →∞, →∞, n, n, 1, vn, n, 1+, n, , 1, 12, , is

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Sequences and Series, , 13, , Σvn is divergent by p – series test. (p = 0 < 1), ∴ By comparison test, Σun is divergent, (Students are advised to follow the procedure, given in ex. 1.2.9(a) and (b) to find “ un ” of the given series.), EXAMPLE 4, , Show that 1 +, , 1 1, 1, + + ....... + + ..... is convergent., 1 2, n, , SOLUTION, 1, (neglecting 1st term ), n, 1, 1, 1, =, <, = n −1, 1.2.3......n 1.2.2.2.....n − 1times (2 ), 1 1, 1, Σun < 1 + + 2 + 3 + ......, ∴, 2 2, 2, 1, which is an infinite geometric series with common ratio < 1, 2, 1, ∴, Σ n −1 is convergent. (1.2.3(a)). Hence Σun is convergent., 2, un =, , EXAMPLE 5, , Test for the convergence of the series,, , 1, 1, 1, +, +, + ......., 1.2.3 2.3.4 3.4.5, , SOLUTION, , un, = Lt, n →∞ v, n →∞, n, , n3, = 1 (finite), 1 ⎞⎛ 2 ⎞, 3⎛, n ⎜ 1 + ⎟⎜ 1 + ⎟, ⎝ n ⎠⎝ n ⎠, ∴ By comparison test, Σun , and Σvn converge or diverge together. But by p-series test,, 1, Σvn = Σ 3 is convergent ( p = 3 > 1 ); ∴ Σun is convergent ., n, , un =, , 1, ;, n ( n + 1)( n + 2 ), , Take vn =, , 1, n3, , Lt, , EXAMPLE 6, , If un = n 4 + 1 − n 4 − 1, show that Σun is convergent., SOLUTION, 1, , 1, , 1 ⎞2, 1 ⎞2, ⎛, ⎛, un = n 2 ⎜ 1 + 4 ⎟ − n 2 ⎜ 1 − 4 ⎟, ⎝ n ⎠, ⎝ n ⎠, , [JNTU, 2005]

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Engineering Mathematics - I, , 14, , ⎡⎛, 1, 1, 1, 1, 1, 1, ⎞ ⎛, ⎞⎤, = n 2 ⎢⎜ 1 + 4 − 8 +, − .... ⎟ − ⎜1 − 4 − 8 −, − ..... ⎟ ⎥, 12, 12, ⎠ ⎝ 2n 8n 16n, ⎠⎦, ⎣⎝ 2n 8n 16n, 1, 1, ⎡1, ⎤ 1 ⎡, ⎤, = n 2 ⎢ 4 + 12 + ....⎥ = 2 ⎢1 + 10 + ....⎥, ⎣ n 8n, ⎦ n ⎣ 8n, ⎦, u, 1, vn = 2 , hence Lt n = 1, n, →∞, n, vn, , Take, , ∴ By comparison test, Σun and Σvn converge or diverge together. But Σvn =, convergent by p –series test (p = 2 > 1) ∴ Σun is convergent., EXAMPLE 7, , Test the series, , 1, 1, 1, +, +, + ..... for convergence., 1+ x 2 + x 3 + x, , SOLUTION, un =, , 1, ;, n+ x, , take vn =, , 1, ,, n, , then, , un, 1, n, =, =, vn n + x 1 + x, n, , ⎛, ⎞, ⎜ 1 ⎟, 1, Lt ⎜, = 1; Σvn = Σ is divergent by p-series test (p =1 ), ⎟, n →∞, n, ⎜1+ x ⎟, n⎠, ⎝, ∴ By comparison test, Σun is divergent., EXAMPLE 8, ∞, ⎛1⎞, Show that Σ sin ⎜ ⎟ is divergent., n =1, ⎝n⎠, SOLUTION, 1, ⎛1⎞, take vn =, un = sin ⎜ ⎟ ;, n, ⎝n⎠, ⎛1⎞, sin ⎜ ⎟, u, ⎝ n ⎠ = Lt sin t (where t = 1 ) = 1, Lt n = Lt, n, n →∞ v, n →∞ ⎛ 1 ⎞, t →0 t, n, ⎜ ⎟, ⎝n⎠, , ∴ Σu , Σvn both converge or diverge . But Σvn = Σ, n, , ( p -series test, p = 1 ); ∴ Σun is divergent., , 1, is divergent, n, , 1, is, n2

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Sequences and Series, , 15, , EXAMPLE 9, , ⎛1⎞, Test the series Σ sin −1 ⎜ ⎟ for convergence., ⎝n⎠, SOLUTION, 1, 1, un = sin −1 ;, Take, vn =, n, n, ⎛1⎞, −1⎜ ⎟, , 1, u, ⎛ θ ⎞ ⎛, ⎞, sin ⎝ n ⎠ ; =, Lt ⎜, = 1⎜ Taking sin −1 = θ ⎟, Lt n = Lt, ⎟, θ, →, 0, n →∞ v, n →∞ ⎛ 1 ⎞, n, ⎝ sin θ ⎠ ⎝, ⎠, n, ⎜ ⎟, n, ⎝ ⎠, , But Σvn is divergent. Hence Σun is divergent., EXAMPLE 10, , Show that the series 1 +, , 1 22 33, + + + ..... is divergent., 22 33 43, , SOLUTION, 1 2 2 33, +, +, + ..... . Therefore, 2 2 33 4 4, nn, nn, 1, ;, =, =, n=, n, n, +, +, n, n, 1, 1, ( )( ), ⎛ 1⎞ n⎛ 1⎞, ⎛ 1 ⎞⎛ 1 ⎞, n ⎜ 1 + ⎟⎜1 + ⎟, n ⎜1 + ⎟ .n ⎜1 + ⎟, ⎝ n⎠ ⎝ n⎠, ⎝ n ⎠⎝ n ⎠, , Neglecting the first term, the series is, un =, , nn, , ( n + 1), , Take vn =, , n +1, , 1, n, , un, 1, 1, 1, = Lt, = Lt, =, n, n, n, n, →∞, →∞, vn, e, ⎛ 1 ⎞⎛ 1 ⎞, ⎛ 1⎞, ⎜ 1 + ⎟⎜1 + ⎟, ⎜1 + ⎟ .1, ⎝ n ⎠⎝ n ⎠, ⎝ n⎠, 1, which is finite and Σvn = Σ, is divergent by p –series test ( p = 1), n, ∴, Σun is divergent., , ∴, , Lt, , n →∞, , EXAMPLE 11, , Show that the series, , 1, 3, 5, +, +, + .......∞ is convergent., 1.2.3 2.3.4 3.4.5, , SOLUTION, , 1, 3, 5, +, +, + .......∞, 1.2.3 2.3.4 3.4.5, , (JNTU 2000)

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Engineering Mathematics - I, , 16, , 1⎞, ⎛, 2− ⎟, ⎜, 2n − 1, 1, n⎠, ⎝, = 2., nth term = un =, n ( n + 1)( n + 2 ) n ⎛ 1 ⎞ ⎛ 2 ⎞, ⎜1 + ⎟ ⎜1 + ⎟, ⎝ n ⎠⎝ n ⎠, 1, vn = 2, Take, n, 1⎞, ⎛, 2− ⎟, ⎜, u, 1, n⎠, ⎛ 1 ⎞, ⎝, Lt n = Lt 2, ÷⎜ 2 ⎟, n →∞ v, n →∞ n, ⎝n ⎠, 1+ 1 1+ 2, n, n, n, u, 2−0, Lt n =, = 2 which is finite and non-zero, n →∞ v, (1 + 0 )(1 + 0 ), n, , (, , But, , ∑v, , n, , 1, , ∑n, , =, , ), , ∑ v converge or diverge together, is convergent. ∴ ∑ u is also convergent., , ∴ By comparison test, , ∑u, , )(, , n, , and, , n, , n, , 2, , EXAMPLE 12, ∞, , Test whether the series, , ∑, n =1, , 1, n + n +1, , is convergent, , (JNTU 1997, 1999, 2003), , SOLUTION, ∞, , ∑, , The given series is, , 1, , n + n +1, 1, un =, n + n +1, n =1, , =, , (, , n +1 − n, n + n +1, , )(, , n +1 − n, , ), , = n +1 − n, , 1, ⎧⎛, 1, 1, ⎪⎧⎛ 1 ⎞ 2 ⎪⎫, ⎞ ⎫, un = n ⎨⎜1 + ⎟ − 1⎬ = n ⎨⎜1 +, − 2 + ..... ⎟ − 1⎬, n⎠, ⎠ ⎭, ⎩⎝ 2n 8n, ⎩⎪⎝, ⎭⎪, , 1, ⎧1, ⎫ 1 ⎧1 1, ⎫, u n = n ⎨ – 2 + ...⎬ =, + ....⎬, ⎨ –, n ⎩ 2 8n, ⎩ 2 n 8n, ⎭, ⎭

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Sequences and Series, , 1, n, , vn =, , Take, , 17, , un, 1 ⎧1 2, ⎫ ⎛ 1 ⎞ 1, = Lt, − + ......⎬ ÷ ⎜, ⎨, ⎟=, n →∞ v, n →∞, n ⎩ 2 8n, ⎭ ⎝ n⎠ 2, n, Lt, , which is finite and non-zero ., un and, Using comparison test, , ∑, 1, ∑v = ∑ n, , But, , n, , ∑u, , ∴, , ∑v, , n, , converge or diverge together., , (, , is divergent since p = 1, , 2, , ), , is also divergent., , n, , EXAMPLE 13, ∞, , ∑ ⎡⎣, , Test for convergence, , n =1, , n3 + 1 − n ⎤, ⎦, , [JNTU 1996, 2003, 2003], , (, , ), , 1, ⎡, ⎤, 1 1 −1, ⎡⎛, 1⎞3 ⎤, 1, 1, 3, 3, ⎢, . 6 + ..... − 1⎥, un = n ⎢⎜ 1 + 3 ⎟ − 1⎥ = n 1 + 3 +, ⎢, ⎥, 3, 1.2, n, n, n, ⎠, ⎢⎣⎝, ⎥⎦, ⎣, ⎦, , th, , n term, , Lt, , Then, , un, , n →∞, , =, , 1, 1, 1 ⎛1 1, 1, ⎞, − 5 + ...... = 2 ⎜ − 3 + ...... ⎟ ; Let vn = 2, 2, n, 3n 9n, n ⎝ 3 9n, ⎠, , vn, , = Lt 1 − 1 3 + .... = 1 ≠ 0, 3, 3, n →∞, 9n, , ∑v, , n, , ), , (, , ∴ By comparison test,, But, , 3, , ∑u, , n, , and, , ∑v, , n, , both converge or diverge., , is convergent by p -series test ( since p = 2 > 1) ∴, , ∑u, , n, , is convergent., , EXAMPLE 14, , Show that the series,, , p≤2, , 2 3, 4, + p + p + ....... is convergent for p > 2 and divergent for, p, 1 2, 3, , SOLUTION, , nth term of the given series = un =, Let us take vn =, , 1, n, , p −1, , ; Lt, , n →∞, , un, , vn, , (, , ) (, , 1, 1+ 1, n +1 n 1+ n, n, =, =, p, p, p −1, n, n, n, , = 1 ≠ 0;, , )

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Engineering Mathematics - I, , 18, , ∑ u and ∑ v, But ∑ v = ∑ 1, n, , ∴, , n, , n, , n, , both converge or diverge by comparison test., converges when p -1>1 ; i.e., p >2 and diverges when, , p −1, , p − 1 ≤ 1 i.e p ≤ 2 ; Hence the result., , EXAMPLE 15, , ⎛ 2n + 3 ⎞, Test for convergence ∑ ⎜ n, ⎟, n =1 ⎝ 3 + 1 ⎠, ∞, , SOLUTION, , (, (, , ⎡ 2n 1 + 3, ⎢, 2n, un = ⎢, n, 1, ⎢⎣ 3 1 + 3n, , ), ), , 1, , 2, , (JNTU 2003), , 1, , ⎤ 2, ⎥, ⎥ ;, ⎥⎦, , 2, vn = n ;, 3, , Take, , un, = 1 ≠ 0 ; ∴ By comparison test,, n →∞ v, n, , ∑u, , Lt, , n, , ∞, , ⎛, un ⎜ 1 +, =, vn ⎜ 1 +, ⎝, , n, , n, , and, , ∑v, , n, , ⎞, 2 ⎟, 1 n⎟, 3 ⎠, , 3, , 1, , 2, , n, , behave the same way., , 3, , 2 2 ⎛2⎞ 2, ⎛2⎞ 2, + + ⎜ ⎟ + ....., which is a geometric series with, But ∑ vn = ∑ ⎜ ⎟ =, 3 3 ⎝3⎠, n =1 ⎝ 3 ⎠, common ratio 2 (<1) ∴ ∑ vn is convergent. Hence ∑ un is convergent., 3, EXAMPLE 16, , Test for convergence of the series,, , 1, 4, 9, +, +, + ......, 4.7.10 7.10.13 10.13.16, , (JNTU 2003), , SOLUTION, , and, , ∴, , 4, 7, 10,..............is an A . P;, , tn = 4 + ( n − 1) 3 = 3n + 1, , 7, 10, 13,............is an A . P;, , tn = 7 + ( n − 1) 3 = 3n + 4, , 10 , 13 , 16 ,.............is an A. P;, , tn = 10 + ( n − 1) 3 = 3n + 7, , un =, =, , n2, =, ( 3n + 1)( 3n + 4 )( 3n + 7 ) 3n 1 + 1, , (, , (, , 27 n 1 + 1, , 3n, , )(, , 1, , 1+ 4, , 3n, , )(, , 1+ 7, , 3n, , ), , ;, , n2, , 3n, , ) .3n (1 + 4 3n ) .3n (1 + 7 3n )

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Sequences and Series, , Taking vn =, , 19, , 1, , we get, n, , un, 1, =, ≠ 0 ; ∴ By comparison test, both, n →∞ v, 27, n, Lt, , manner. But by p –series test,, , ∑v, , n, , ∑u, , n, , and, , ∑v, , n, , is divergent, since p = 1. ∴, , behave in the same, , ∑u, , n, , is divergent ., , EXAMPLE 17, , Test for convergence, , 2 n 2 − 5n + 1, ∑ 4n 3 − 7 n 2 + 2, , (JNTU 2003), , SOLUTION, , nth term of the given series = un =, Let, , vn =, , 2 n 2 − 5n + 1, 4n3 − 7 n 2 + 2, , 1, n2, , ⎡ n 2− 5 + 1, ⎤, ⎡ 2− 5 + 1, 2, un, n2 ⎥, n, n, ⎢, ⎢, n2, n, = Lt . ⎢, × ⎥ = Lt ⎢, Lt, n →∞ v, n →∞, n →∞, 3, 1, 7, 7, 2, 2, n, ⎢⎣ n 4 − n + n3, ⎥⎦, ⎢⎣ 4 − n + n3, ∴ By comparison test, ∑ un and ∑ vn both converge or diverge., , ), , (, , But, , ∑v, , n, , ), , (, , is convergent. [p series test – p = 2 > 1] ∴, , ∑u, , n, , ⎤, 2, ⎥, ⎥= 4 ≠0, ⎥⎦, , is convergent ., , EXAMPLE 18, , Test the series, , ∑u, , n, , , whose nth term is, , ( 4n, , 1, 2, , − i), , SOLUTION, , ⎡, ⎤, un, n2, ⎢, ⎥ 1, un =, ;, Lt, = Lt ⎢, 2, ⎥= 4≠0, n →∞ v, n →∞, 2, i, ( 4n − i ), n, ⎢⎣ n 4 − n 2 ⎥⎦, ∴ ∑ un and ∑ vn both converge or diverge by comparison test. But ∑ vn is, , 1, , 1, Let vn = 2 ,, n, , convergent by p –series test ( p = 2 > 1) ; ∴, ∞, , Note: Test the series, , ∑ 4n, n =1, , 1, 2, , –1, , (, , ∑u, , n, , is convergent., , )

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Engineering Mathematics - I, , 20, , EXAMPLE 19, , ⎛1⎞, ⎝n⎠, , ⎛1⎞, ⎝n⎠, , If un = ⎜ ⎟ .sin ⎜ ⎟ , show that, , ∑u, , n, , is convergent ., , SOLUTION, , 1, , so that, n2, , Let vn =, , ⎛u, Lt ⎜ n, n →∞ v, ⎝ n, , ∑v, , is convergent by p –series test ., , n, , ( ), ( ), , sin 1, ⎞, n = Lt ⎛ sin t ⎞, =, Lt, ⎟ n →∞, ⎜, ⎟, t →0, 1, ⎝ t ⎠, ⎠, n, , ⎛ un, ⎝ vn, , where t = 1/n, Thus Lt ⎜, n →∞, , ∑u, , ∴ By comparison test,, , ⎞, ⎟ =1≠ 0, ⎠, n, , is convergent., , EXAMPLE 20, , Test for convergence, , 1, tan( 1 ), n, n, , ∑, , SOLUTION, , Take vn = 1, , n, , 3, , u, ; Lt ⎡ n, n →∞, , 2, , ⎢⎣, , Hence by comparison test,, , ⎤ = 1 ≠ 0 ( as in above example), vn ⎥⎦, , ∑u, , n, , converges as, , ∑v, , n, , converges., , EXAMPLE 21, ∞, , Show that, , ∑ sin, n =1, , 2, , ⎛1⎞, ⎜ ⎟ is convergent., ⎝n⎠, , SOLUTION, , ⎛1⎞, un = sin ⎜ ⎟ ;, ⎝n⎠, 2, , Let, , 1, Take vn = 2 ,, n, , ( ) ⎤⎥, , ⎡ sin 1, ⎛ un ⎞, n, Lt ⎜ ⎟ = Lt ⎢, n →∞ v, n →∞ ⎢, 1, ⎝ n⎠, n, ⎣, , 2, , ⎛ sin t ⎞, = Lt ⎜, ⎟, →, 0, t, ⎥, ⎝ t ⎠, ⎦, , u, t = 1 ; Lt ⎛⎜ n ⎞⎟ = 12 = 1 ≠ 0, n n→∞ ⎝ vn ⎠, ∴ By comparison test, ∑ un and ∑ vn behave the same way., where, , But, , ∑v, , n, , is convergent by p- series test, since p = 2 > 1; ∴, , ∑u, , n, , is convergent., , 2

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Sequences and Series, , 21, , EXAMPLE 22, ∞, , ∑, , Show that, , 1, , n=2, , log ( n n ), , is divergent., , SOLUTION, , un = 1, , ; log 2 < 1 ⇒ 2 log 2 < 2 ⇒ 1, >1 ;, n log n, 2 log 2, 2, 1, > 1 ..... 1, > 1 ,n∈ N, Similarly, 3log 3, 3,, n log n, n, ∴, ∑ 1, > ∑ 1 ; But ∑ 1 is divergent by p-series test., n log n, n, n, By comparison test, given series is divergent. [If ∑ vn is divergent and un ≥ vn ∀n then, , ∑u, , n, , is divergent.], , (Note : This problem can also be done using Cauchy’s integral Test., EXAMPLE 23, ∞, , ∑ (c + n) ( d + n), , Test the convergence of the series, , −r, , −s, , , where c, d, r, s are all +ve., , n =1, , SOLUTION, , The nth term of the series = un =, , vn =, , Let, , Lt, , n →∞, , 1, n, , r +s, , un, = 1 ≠ 0 , ∴ ∑ un and, vn, , ∑v, , But by p-series test,, , ∴, , un, =, vn, , Then, , ∑u, , n, , n, , 1, , (c + n) ( d + n), r, , s, , ., , nr +s, r, , ⎛ c⎞, ⎛ d⎞, n ⎜ 1 + ⎟ .n s ⎜1 + ⎟, ⎝ n⎠, ⎝ n⎠, r, , ∑v, , n, , s, , =, , 1, r, , ⎛ c⎞ ⎛ d⎞, ⎜1 + ⎟ ⎜1 + ⎟, ⎝ n⎠ ⎝ n⎠, , both converge are diverge, by comparison test., , converges if (r + s) > 1 and diverges if (r + s) ≤ 1, , converges if ( r + s ) > 1 and diverges if ( r + s ) ≤ 1., , EXAMPLE 24, ∞, , Show that, , ∑n, , (, , − 1+ 1, , n, , ) is divergent., , 1, , SOLUTION, , un = n, , (, , − 1+ 1, , n, , )= 1, 1, n.n, , s, , Take, n, , vn =, , u, 1, 1, ; Lt n = Lt 1 = 1 ≠ 0, n, n, →∞, →∞, n, vn, n n

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Engineering Mathematics - I, , 22, , Lt, , For let, , 1, 1, = y say; log y = Lt − .log n = − Lt n = 0, n →∞ n, n →∞ 1, , 1, , n →∞, , n, , 1, , n, , ⎛∞⎞, ⎟ using L Hospitals rule), ⎝∞⎠, converge or diverge. But p-series test,, By comparison test both ∑ un and ∑ vn, , ∴, , y = e0 = 1, , ∑v, , n, , (⎜, , diverges (since p =1); Hence, , ∑u, , n, , diverges., , EXAMPLE 25, , (n + a), ∑, p, q, n =1 ( n + b ) ( n + c ), r, , ∞, , Test for convergence the series, SOLUTION, , un =, Take vn =, , (n + a ), (n + b ) p (n + c )q, r, , 1, n, , p+ q −r, , =, , (, , np, , n, , q, , q, , un, =1≠ 0 ;, n →∞ v, n, , 1, , =, , n, , p+ q −r, , ., , (, , (, , 1+ b, , )(, p, , n, , n, , ), , r, , 1+ c, , n, , ), , q, , ;, , ; Lt, , But by p-series test,, , ∑u, , p, , 1+ a, , r, , nr 1 + a, , Applying comparison tests both, , Hence, , ), n, (1 + b n ) n (1 + c n ), , , a, b, c , p, q, r, being +ve., , ∑v, , n, , ∑u, , n, , and, , ∑v, , n, , converge or diverge., , converges if ( p+ q –r ) > 1and diverges if ( p + q –r) ≤ 1., , converges if ( p + q – r ) > 1 and diverges if ( p + q – r) ≤ 1., , EXAMPLE 26, Test the convergence of the following series whose nth terms are:, , ( 3n + 4 ), ( 2n + 1)( 2n + 3)( 2n + 5), , (a), (d), , 1, ;, ( 3 + 5n ), n, , ;, , 1, ;, n, , (b), , tan, , (e), , 1, n.3n, , (c), , ⎛ 1 ⎞ ⎛ n +1 ⎞, ⎜ 2 ⎟⎜, ⎟, ⎝ n ⎠⎝ n + 3 ⎠, , SOLUTION, , (a), , Hint : Take, , vn =, , ⎛u ⎞ 3, 1, ; v is convergent; Lt ⎜ n ⎟ = ≠ 0 (Verify), 2 ∑ n, n →∞ v, n, ⎝ n⎠ 8, , Apply comparison test:, un is convergent [the student is advised to work out this problem fully ], , ∑, , n

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Sequences and Series, , (b), , 23, , Proceed as in Example 8;, , ∑u, , n, , is convergent., , ⎛u ⎞, (1 + 1n ) = e = 1 ≠ 0, 1, Hint : Take vn = 2 ; Lt ⎜ n ⎟ = Lt, n, n →∞ v, n, e3 e 2, ⎝ n ⎠ n→∞ 1 + 3, n, 1, vn = 2 is convergent (work out completely for yourself ), n, n, , (c), , (d), , (, , un =, , ∑u, , n, , ⎛u ⎞, 1, 1, 1, 1, ; Take vn = n ; Lt ⎜ n ⎟ = 1 ≠ 0, = n., n, n, n →∞ v, 5, 3 +5, 5 ⎡ ⎛3⎞ ⎤, ⎝ n⎠, ⎢1 + ⎜ ⎟ ⎥, ⎣⎢ ⎝ 5 ⎠ ⎦⎥, n, , and, , ∑v, , n, , behave the same way. But, , geometric series with common ratio, , ∴, (e), , ), , ∑u, , n, , ∑v, , n, , is convergent since it is a, , 1, <1, 5, , is convergent by comparison test ., , 1, 1, ≤ n , ∀n ∈ N , since, n, n.3, 3, 1, 1, ∴, ∑ n.3n ≤ ∑ 3n, , n.3n ≥ 3n ;, …..(1), , The series on the R.H .S of (1) is convergent since it is geometric series with, , r=, , 1, <1., 3, , ∴ By comparison test, , 1, , ∑ n.3, , n, , is convergent., , EXAMPLE 27, , Test the convergence of the following series., , 1+ 2, 1+ 2 + 3, 1+ 2 + 3 + 4, + ..............., + 2, + 2, 2, 2, 2, 2, 1 + 2 1 + 2 + 3 1 + 22 + 32 + 42, , (a), , 1+, , (b), , 12 + 22 12 + 22 + 32 12 + 22 + 32 + 42, 1 + 3 3 + 3 3 3 + 3 3 3 3 + .............., 1 +2 1 +2 +3 1 +2 +3 +4

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Engineering Mathematics - I, , 24, , SOLUTION, , n, , 1 + 2 + 3 + .... + n, un = 2, =, 1 + 22 + 32 + .....n 2, , ( n + 1), , 3, 2, =, ( 2n + 1) ( 2n + 1), n ( n + 1), 6, un, 1, n, 3, ⎛, ⎞ 3, Take vn =, ; Lt, = Lt ⎜, = ≠0, ⎟, n →∞ v, n →∞ 2n + 1, n, ⎝, ⎠ 2, n, , (a), , ∑u, , (b), , and, , n, , ∑v, , n, , behave alike by comparison test., , But, , ∑v, , un =, , 1 + 2 + .... + n, =, 13 + 23 + .....n3, , n, , 2, , is diverges by p-series test. Hence, 2, , 2, , n ( n + 1), n2, , ( 2n + 1), 6, , ( n + 1), , 2, , =, , ∑u, , n, , is divergent., , 2 ( 2n + 1), 3n ( n + 1), , 4, 1, Hint : Take vn = and proceed as in (a) and show that, n, , ∑u, , n, , is divergent., , Exercise 1.1, 1. Test for convergence the infinite series whose nth term is:, , (a), , 1, n− n, , [Ans : divergent], , (b), , n +1 − n, n, , [Ans : convergent], , (c), , n2 + 1 − n, , [Ans : divergent], , (d), (e), (f), , (g), (h), , n, n −1, 2, , n3 + 1 − n3, 1, n ( n + 1), , n, n +1, 2n 3 + 5, 4n 5 + 1, 2, , [Ans : convergent], [Ans : divergent], [Ans : divergent], , [Ans : convergent], [Ans : convergent]

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Sequences and Series, , 25, , 2. Determine whether the following series are convergent or divergent., , 1, 2, 3, +, +, + ............, −1, −2, 1+ 3, 1+ 3, 1 + 3−3, 12 22 32, 2 + 10n, + 3 + 3 + ........ +, + ..... ......., 3, 1, 2, 3, n3, 1, 1, 1, +, +, + ...... ........, 1+ 2, 2+ 3, 3+ 4, , (a), (b), (c), , 2 3 4, + + + ...... ........., 32 42 52, 1 1 1, + + + .........., 12 23 34, , (d), (e), , ∞, , ∑, , (f), , n =1, , 4, , 4n 2 + 2 n + 3, , ∑ (8, ∞, , (g), , n2 + 1, , 3, , 1, , n, , [Ans : convergent], [Ans : divergent], [Ans : divergent], [Ans : convergent], , ..............., , [Ans : divergent], , ), , − 1 ...................., , 1, , [Ans : divergent], , [Ans : divergent], , (h), , 3n3 + 8, ∑1 5n5 + 9 ........................, , [Ans : convergent], , (i), , 1, 2, 3, +, +, + ..........., 1.3 3.5 5.7, , [Ans : divergent], , ∞, , 1.3.3 D’ Alembert’s Ratio Test, , Let (i), , ∑u, , n, , be a series of +ve terms and (ii) Lt, , Then the series, , n →∞, , ∑u, , n, , un +1, = k ( ≥ 0), un, , is (i) convergent if k < 1 and (ii) divergent if k > 1., , Proof :, , un +1, = k ( < 1), n →∞ u, n, , Case (i) Lt, , From the definition of a limit, it follows that, , ∃m > 0 and l ( 0 < l < 1) ∋, , un +1, < l∀n ≥ m, un

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Engineering Mathematics - I, , 26, , um +1, u, < l , m + 2 < l ,.........., um, um +1, , i.e.,, , ∴, , ⎡ u, ⎤, u, um + um +1 + um + 2 + ...... ........ = um ⎢1 + m +1 + m + 2 + .....⎥, um, um, ⎣, ⎦, um 1, , um 1, um, , um, um, , < um (1 + l + l 2 + ...) = um ., But um ., , 2, 1, , ., , um 1, um, , ....., , 1, ( l < 1), 1− l, , 1, is a finite quantity ∴, 1− l, , ∞, , ∑u, , n=m, , n, , is convergent, , By adding a finite number of terms u1 + u2 + ...... + um −1 , the convergence of the, ∞, , series is unaltered., , ∑u, , n=m, , n, , is convergent., , un +1, = k >1, n →∞ u, n, , Case (ii) Lt, , There may be some finite number of terms in the beginning which do not satisfy, the condition, , ∋, , un +1, ≥ 1 . In such a case we can find a number ‘m’, un, , un +1, ≥ 1, ∀n ≥ m, un, , Omitting the first ‘m’ terms, if we write the series as u1 + u2 + u3 + ........., we, have, , u, u2, u, ≥ 1, 3 ≥ 1, 4 ≥ 1 .......... and so on, u1, u2, u3, ∴, , ⎛ u u u, ⎞, u1 + u2 + ...... + un = u1 ⎜1 + 2 + 3 . 2 + ..... ⎟ (to n terms), ⎝ u1 u2 u1, ⎠, , ≥ u1 (1 + 1 + 1.1 +.......to n terms), = nu1, Lt, , n →∞, , n, , ∑u, n =1, , n, , ≥ Lt n.u1 which → ∞ ; ∴, n →∞, , ∑u, , n, , is divergent .

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Sequences and Series, , 27, ∞, , Note: 1 The ratio test fails when k = 1. As an example, consider the series,, , 1, , ∑n, n =1, , p, , p, , p, ⎛ 1 ⎞, un +1, ⎛ n ⎞, ⎜, ⎟ =1, = Lt ⎜, =, Lt, Lt, ⎟, n →∞ u, n →∞ n + 1, n →∞ ⎜, 1+ 1 ⎟, ⎝, ⎠, n, n⎠, ⎝, , Here, , i.e.,, k = 1 for all values of p,, But the series is convergent if p > 1 and divergent if p ≤ 1 , which shows that, when k = 1, the series may converge or diverge and hence the test fails ., Ratio test can also be stated as follows:, , Note: 2, , If, , ∑u, , n, , un, = k , then, n →∞ u, n +1, , is series of +ve terms and if Lt, , ∑u, , n, , is convergent, , If k > 1 and divergent if k < 1 (the test fails when k = 1)., , Solved Examples, Test for convergence of Series, EXAMPLE 28, , x, x2, x3, +, +, + ................., 1.2 2.3 3.4, , (a), , SOLUTION, , un =, , xn, x n +1, ; un +1 =, ;, n ( n + 1), ( n + 1)( n + 2 ), un +1, =x, n →∞ u, n, , ∴ By ratio test, , ∑u, , When x = 1, un =, , n, , n, , is convergent When |x| < 1 and divergent when | x | > 1;, , u, 1, 1, ; Take vn = 2 ; Lt n = 1, n (1 + 1 n ), n n→∞ vn, 2, , ∴ By comparison test, , ∑u, , n ( n + 1), un +1, x n +1, 1, =, =, x., ., n, un, ⎛ 2⎞, ( n + 1)( n + 2 ) x, ⎜1 + ⎟, ⎝ n⎠, , Lt, , Therefore, , Hence, , (JNTU 2003), , ∑u, , n, , is convergent., , is convergent when x ≤ 1 and divergent when x > 1 .

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Engineering Mathematics - I, , 28, , 1 + 3x + 5 x 2 + 7 x 3 + ......., , (b), , SOLUTION, , un = ( 2n − 1) x n −1 ;, , ∴ By ratio test, , ∑u, , n, , ∑u, , Lt, , is convergent when x < 1 and divergent when x > 1, , When x = 1: un = 2n − 1; Lt un = ∞ ; ∴, Hence, , un +1, ⎛ 2n + 1 ⎞, = Lt ⎜, ⎟x = x, n →∞ u, n →∞ 2n − 1, ⎝, ⎠, n, , un +1 = ( 2n + 1) x n ;, , n →∞, , ∑u, , n, , is divergent., , is convergent when x < 1 and divergent when x ≥ 1, , n, , ∞, , (c), , xn, ..........., ∑, 2, n =1 n + 1, , SOLUTION, , un =, , xn, ;, n2 + 1, , un +1 =, , x n +1, , ( n + 1), , 2, , +1, , ., , (, , ), , ⎡ 2, ⎤, 1 2 ⎥, n, 1, +, ⎢, un +1, un +1 ⎛ n + 1 ⎞, n, ⎥ ( x) = x, Hence, = Lt ⎢, =⎜ 2, ⎟ x , nLt, →∞ u, n →∞, 2, un ⎝ n + 2n + 2 ⎠, ⎛, ⎢ n2 1 + + 2 ⎞ ⎥, n, ⎢⎣ ⎜⎝ n n 2 ⎟⎠ ⎥⎦, ∴ By ratio test, ∑ un is convergent when x < 1 and divergent when x > 1 When, 2, , 1, 1, ; Take vn = 2, n +1, n, ∴ By comparison test, ∑ un is convergent when x ≤ 1 and divergent when x > 1, , x = 1: un =, , 2, , EXAMPLE 29, , ⎛ n2 − 1 ⎞ n, ∑ ⎜ 2 ⎟x , x > 0 for convergence., n →∞ ⎝ n + 1 ⎠, ∞, , Test the series, SOLUTION, , ⎡ ( n + 1)2 − 1 ⎤ n +1, ⎛ n2 − 1 ⎞ n, un = ⎜ 2 ⎟ x ; un +1 = ⎢, ⎥x, 2, ⎢⎣ ( n + 1) + 1 ⎥⎦, ⎝ n +1 ⎠

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Sequences and Series, , 29, , ⎡⎛ n 2 + 2n ⎞ ( n 2 + 1) ⎤, un +1, ⎥ .x, = Lt ⎢⎜ 2, Lt, ⎟ 2, n →∞ u, n →∞, ⎢⎣⎝ n + 2n + 2 ⎠ ( n − 1) ⎥⎦, n, , (, , (, , ∴ By ratio test,, , ∑, , ), , ⎡, ⎤, n 4 (1 + 2 n ) 1 + 1 n 2, ⎥=x, = Lt ⎢ 4, 2, n →∞ ⎢ n 1 + 2 n + 2 n 2, n, 1, –, 1, ⎥⎦, ⎣, un is convergent when x < 1 and divergent when x > 1 when x = 1,, , )(, , ), , n2 − 1, 1, Take vn = 0, 2, n +1, n, , un =, , Applying p-series and comparison test, it can be seen that, , ∴, , ∑u, , n, , ∑u, , n, , is divergent when x = 1., , is convergent when x < 1 and divergent x ≥ 1, , EXAMPLE 30, , Show that the series 1 +, , 2 p 3p 4 p, + +, + ..... , is convergent for all values of p., 2, 3, 4, , SOLUTION, , ( n + 1), np, un =, ; un +1 =, n, n +1, , p, , ⎡ ( n + 1) p n ⎤, ⎧⎪ 1 ⎛ n + 1 ⎞ p ⎫⎪, un +1, Lt, = Lt ⎢, × p ⎥ = Lt ⎨, ⎜, ⎟ ⎬, n →∞ u, n →∞, ⎢⎣ n + 1 n ⎥⎦ n→∞ ⎩⎪ ( n + 1) ⎝ n ⎠ ⎭⎪, n, p, , ∑u, , 1, ⎛ 1⎞, × Lt ⎜1 + ⎟ = 0 < 1 ;, = Lt, n →∞ ( n + 1), n →∞, ⎝ n⎠, n, , is convergent for all ‘ p ‘ ., , EXAMPLE 31, , Test the convergence of the following series, , 1 1, 1, 1, + p + p + p + ............, p, 1 3 5, 7, SOLUTION, , un =, , 1, , ( 2n − 1), , p, , ;, , un +1 =, , 1, , ( 2n + 1), , p

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Engineering Mathematics - I, , 30, , 2 p.n p (1 − 1 2n ), un +1 ( 2n − 1), =, =, ;, p, p, un, ( 2n + 1) 2 p n p (1 + 1 2n ), p, , p, , ∴ Ratio test fails., 1 u, np, Take vn = p ; n =, =, n vn ( 2n − 1) p, which is non – zero and finite, ∴ By comparison test,, un and, , ∑, But by p – series test, ∑ v, , n, , when p ≤ 1, , ∴, , ∑u, , n, , =, , 1, 1 ⎞, ⎛, 2 p ⎜1 − ⎟, ⎝ 2n ⎠, , ∑v, , n, , 1, , ∑n, , p, , p, , un +1, =1, n →∞ u, n, Lt, , un, 1, = p,, n →∞ v, 2, n, , ; Lt, , both converge or both diverge., , converges when p > 1 and diverges, , is convergent if p > 1 and divergent if p ≤ 1 ., , EXAMPLE 32, ∞, , Test the convergence of the series, , ∑, n =1, , SOLUTION, , un =, , (n + 1)x n ; x > 0, n3, , ( n + 1) x n ; u ( n + 2 ) x n+1, n +1, 3, n3, ( n + 1), , un +1, n + 2 n +1, n3, ⎛ n + 2 ⎞⎛ n ⎞, =, ., ., =⎜, .x, x, 3, n, un, ( n + 1) x ⎝ n + 1 ⎟⎠ ⎜⎝ n + 1 ⎟⎠, ( n + 1), 3, , 2⎞, ⎛, ⎜ 1+ n ⎟, un +1, 1, = Lt ⎜, .x = x, Lt, ⎟, 3, n →∞ u, n →∞, 1, 1, ⎛, ⎞, n, ⎜ 1+ ⎟ 1+, ⎝ n ⎠ ⎜⎝ n ⎟⎠, , ∴ By ratio test,, , ∑u, , When x = 1, un =, , n, , converges when x < 1 and diverges when x > 1 ., , n +1, n3, , 1, ; By comparison test ∑ un is convergent ( give proof ), n2, ∴ ∑ un is convergent if x ≤ 1 and divergent if x > 1., Take vn =

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Sequences and Series, , 31, , EXAMPLE 33, , Test the convergence of the series, ∞, , (i), , ⎛n, , 2, , ⎝, , n, , ∑⎜ 2, n −1, , +, , (JNTU 2002), , 2.5.8 2.5.8.11, 1 1.2 1.2.3, 1 ⎞, (ii) 1 +, +, +, +, + ... (iii) +, 2 ⎟, 1.5.9 1.5.9.13, 3 3.5 3.5.7, n ⎠, , SOLUTION, , ⎛ n2 1 ⎞, ∑, ⎜ n+ 2⎟ =, n ⎠, n −1 ⎝ 2, ∞, , (i), , un +1, , ( n + 1), =, , 2, , 2n +1, , ∞, , n2 ∞ 1, +∑ 2, ∑, n, n =1 2, n =1 n, , Let un =, , ( n + 1) . 2n, u, ; n +1 =, un, 2n +1 n 2, 2, , n2, 1, ; vn = 2, n, 2, n, 2, , u, 1 ⎛ 1⎞ 1, Lt n +1 = Lt . ⎜1 + ⎟ = < 1, n →∞ u, n →∞ 2, 2, ⎝ n⎠, n, , ∑ u is convergent. By p –series test, ∑ v, ∴ The given series ( ∑ u + ∑ v ) is convergent., ∴ By ratio test, , n, , n, , n, , (ii), , is convergent., , n, , Neglecting the first term, the series can be taken as,, , 2.5.8 2.5.8.11, +, +, 1.5.9 1.5.9.13, , Here, 1st term has 3 fractions ,2nd term has 4 fractions and so on ., , ∴ nth term contains ( n + 2 ) fractions, 2. 5. 8.......are in A. P., , ∴ ( n + 2), , th, , term = 2 + ( n + 1 ) 3 = 3n + 5 ;, , ∴ 1. 5. 9,.......are in A. P., th, ∴ ( n + 2 ) term = 1 + ( n + 1 ) 4 = 4n + 5, ∴, , un =, , 2.5.8..... ( 3n + 5 ), 1.5.9..... ( 4n + 5 ), , un +1 =, un +1, un, , 2.5.8..... ( 3n + 5 )( 3n + 8 ), 1.5.9..... ( 4n + 5 )( 4n + 9 ), , ( 3n + 8) ;, =, ( 4n + 9 ), , ∴ By ratio test,, , ∑u, , n, , 8⎞, ⎛, n⎜3+ ⎟, u, n⎠ 3, = <1, Lt n +1 = Lt ⎝, n →∞ u, n →∞ ⎛, 9⎞ 4, n, n⎜4 + ⎟, n⎠, ⎝, is convergent.

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Engineering Mathematics - I, , 32, , (iii), , 1, 2, 3, ........ are in A. P nth term = n ; 3. 5. 7..........are in A.P. nth term = 2n + 1, , ⎡ 1.2.3.....n ⎤, un = ⎢, ⎥, ⎣ 3.5.7..... ( 2n + 1) ⎦, ⎡, ⎤, 1.2.3.....n ( n + 1), un +1 = ⎢, ⎥, ⎣ 3.5.7..... ( 2n + 1)( 2n + 3) ⎦, , ∴, , un +1 ⎛ n + 1 ⎞, =⎜, ⎟, un ⎝ 2n + 3 ⎠, , ∴, , ⎛ 1⎞, n. ⎜1 + ⎟, u, n⎠ 1, = <1, Lt n +1 = Lt ⎝, n →∞ u, n →∞ ⎛, 3⎞ 2, n, n⎜2 + ⎟, n⎠, ⎝, By ratio test, ∑ un is convergent., , EXAMPLE 34, , 1.3.5.... ( 2n − 1) n −1, .x ( x > 0 ), 2.4.6....2n, n =1, ∞, , Test for convergence, , ∑, , SOLUTION, , The given series of +ve terms has un =, and, , un +1 =, , 1.3.5.... ( 2n + 1) n, x, 2.4.6.... ( 2n + 2 ), , (JNTU 2001), , 1.3.5.... ( 2n − 1) n −1, .x, 2.4.6....2n, , (, (, , ), ), , 2n 1 + 1, un +1, ⎛ 2n + 1 ⎞, 2n .x = x, x, = Lt ⎜, =, Lt, ⎟, n →∞ u, n →∞ 2n + 2, n →∞, ⎝, ⎠, 2n 1 + 2, n, 2n, ∴ By ratio test, ∑ un is converges when x < 1 and diverges when x > 1 when x = 1, the, Lt, , test fails., , 1.3.5.... ( 2n − 1), < 1 and Lt un ≠ 0, n →∞, 2.4.6.....2n, ∴ ∑ un is divergent. Hence ∑ un is convergent when x < 1, and divergent when x ≥ 1, Then un =

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Sequences and Series, , 33, , EXAMPLE 35, , ⎛ 2n − 2 ⎞ n −1, 2, 6 2, Test for the convergence of 1 + x + x + ........ + ⎜ n, ⎟ x + ..... ( x > 0 ), 5, 9, ⎝ 2 +1 ⎠, (JNTU 2003), SOLUTION, , ⎛ 2n − 2 ⎞ n −1, ⎟ x , ( n ≥ 2 ) and ' un ' are all +ve., n, ⎝ 2 +1 ⎠, , Omitting 1st term, un = ⎜, , (2, =, (2, , n +1, , un +1, , − 2), , ⎛ un +1 ⎞, ⎛ 2n +1 − 2 ⎞ ⎛ 2n + 1 ⎞, n, x, ;, =, Lt, Lt, ., ⎜, ⎟, ⎜ n +1, ⎟×⎜ n, ⎟ .x, n +1, n →∞, + 1), ⎝ un ⎠ n→∞ ⎝ 2 + 1 ⎠ ⎝ 2 − 2 ⎠, , ) (, ) (, , (, (, , ), ), , ⎡ 2n +1 1 − 1, 2n 1 + 1 n ⎤, n, ⎢, 2, 2 .x ⎥ = x ;, = Lt, ., ⎥, n →∞ ⎢ n +1, n, 1, 2, ⎢⎣ 2 1 + 2n +1 2 1 − 2n ⎥⎦, Hence, by ratio test, ∑ un converges if x < 1 and diverges if x > 1., When x = 1, the test fails. Then un =, Hence, , ∑u, , n, , 2n − 2, ; Lt un = 1 ≠ 0 ; ∴, 2n + 1 n→∞, , ∑u, , n, , diverges, , is convergent when x < 1 and divergent x > 1, , EXAMPLE 36, ∞, , Using ratio test show that the series, , ∑, , ( 3 − 4i ), n!, , n=0, , SOLUTION, , un, , ( 3 − 4i ), =, , n, , n!, , Hence, by ratio test,, , ;, , ∑u, , un +1, n, , ( 3 − 4i ), =, , n, , converges, , n +1, , (n + 1)!, , (JNTU 2000), , ⎛ un +1 ⎞, ⎛ 3 − 4i ⎞, ⎟ = nLt, ⎜, ⎟ = 0 <1, ⎝ un ⎠ →∞ ⎝ n + 1 ⎠, , ; Lt ⎜, n →∞, , converges., , EXAMPLE 37, , Discuss the nature of the series,, , 2, 3 2 4 3, x+, x +, x + ........∞ ( x > 0 ), 3.4, 4.5, 5.6, , SOLUTION, , Since x > 0 , the series is of +ve terms ;, , (JNTU 2003)

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Engineering Mathematics - I, , 34, , un =, , ( n + 1) x n > u = ( n + 2 ) x n+1, n +1, ( n + 2 )( n + 3), ( n + 3)( n + 4 ), , ⎡, 2, 2, 2, un +1 ⎡ ( n + 2 ) .x ⎤, ⎢ n (1 + 2 n ) .x, =⎢, Lt, ⎥ = Lt ⎢, n →∞ u, +, +, 1, 4, n, n, (, )(, ), ⎢, ⎥⎦ n→∞ ⎢ n 2 1 + 5 + 4 2, n, ⎣, n, n, ⎣, Therefore by ratio test, ∑ un converges if x < 1 and diverges if x >1, , (, , When x = 1, the test fails; Then un =, Taking vn =, , u, 1, ; Lt = n = 1 ≠ 0, →∞, n, n, vn, , ∴ By comparison test, , ∑u, , n, , and, , ), , ⎤, ⎥, ⎥ = x;, ⎥⎦, , ( n + 1) ;, ( n + 2 )( n + 3), , ∑v, , n, , behave same way. But, , ∑v, , n, , is divergent by p-, , series test. (p = 1);, ∴, un is diverges when x = 1, , ∑, ∴ ∑u, , n, , is convergent when x < 1 and divergent when x ≥ 1, , EXAMPLE 38, , Discuss the nature of the series, , 3.6.9.....3n.5n, ∑ 4.7.10.....( 3n + 1)( 3n + 2 ), , SOLUTION, , Here,, , 3.6.9.....3n, 5n, ;, un =, 4.7.10..... ( 3n + 1) ( 3n + 2 ), un +1 =, , 3.6.9.....3n ( 3n + 3) 5n +1, ;, 4.7.10..... ( 3n + 1)( 3n + 4 )( 3n + 5 ), , ( 3n + 2 )( 3n + 3) .5, un +1, = Lt, n →∞ u, n →∞ ( 3n + 4 )( 3n + 5 ), n, Lt, , (, , )(, )(, , ⎡ 5.9n 2 1 + 2, 1+ 3, n, 3, 3n, ⎢, = Lt, 2, n →∞ ⎢, 9n 1 + 4, 1+ 5, 3n, 3n, ⎣, ∴ By ratio test, ∑ un is divergent., , (, , ) ⎤⎥ = 5 > 1, , ) ⎥⎦, , (JNTU 2003)

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Sequences and Series, , 35, , EXAMPLE 39, ∞, , Test for convergence the series, , ∑n, , 1− n, , n =1, , SOLUTION, , un = n1− n ; un +1 = ( n + 1) ;, −n, , u n +1 (n + 1)– n, nn, 1⎛ n ⎞, =, =, = ⎜, ⎟, 1– n, n, un, n ⎝ n +1⎠, n, n(n + 1), , n, , n, , u, 1 ⎛ 1 ⎞⎟, 1, Lt n +1 = Lt . ⎜, = 0. = 0 < 1, n →∞ u, n →∞ n ⎜, e, 1+ 1 ⎟, n, n⎠, ⎝, ∴ By ratio test ∑ un , is convergent, EXAMPLE 40, ∞, , 2n3, Test the series ∑, , for convergence., n =1 n, SOLUTION, , 2 ( n + 1), 2n3, ; un +1 =, un =, n +1, n, , 3, , (, , 3, 2, 1+ 1, un +1 2 ( n + 1), n ( n + 1), n, =, × 3=, =, 3, un, n +1, n, n, 2n, u, Lt n +1 = 0 < 1 ;, n →∞ u, n, , ∴ By ratio test,, , ∑u, , n, , is convergent., , EXAMPLE 41, , Test convergence of the series, SOLUTION, , un =, , 2n n !, ∑ nn, , 2n +1 ( n + 1) !, 2n n !, u, ;, =, ;, n +1, n +1, nn, ( n + 1), , ), , 2, , ;

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Engineering Mathematics - I, , 36, n +1, un +1 2 ( n + 1) ! n n, ⎛ n ⎞, . n = 2⎜, =, ⎟, n +1, un, 2 n!, ⎝ n +1 ⎠, ( n + 1), , un +1, 1, = 2 Lt, n →∞ u, n →∞, n, 1+ 1, Lt, , ∴ By ratio test,, , (, , ∑u, , n, , n2 + 1, 3n + 1, , ∑u, , (b), , 1 + 2n, , (d), , ), , =, , 2, < 1 (since 2 < e < 3), e, , is convergent., , EXAMPLE 42, Test the convergence of the series, , (a), , n, , n, , n, , (e), , 1 + 3n, , n, , where un is, , x n −1, , ( 2n + 1), , a, , , ( a > 0), , (c), , ⎛ 1.2.3....n, ⎞, ⎜, ⎟, ⎝ 4.7.10.....3n + 3 ⎠, , ⎛ 3n3 + 7 n 2 ⎞ n, ⎜, ⎟x, 9, n, 5, 11, +, ⎝, ⎠, , SOLUTION, , (a), , ⎡ ( n + 1)2 + 1 3n + 1 ⎤, ⎛ un +1 ⎞, Lt ⎜, × 2 ⎥, ⎟ = Lt ⎢ n +1, n →∞, n + 1 ⎥⎦, ⎝ un ⎠ n→∞ ⎢⎣ 3 + 1, ⎡ n 2 ⎛1 + 2 + 2 ⎞ 3 n ⎛1 +, ⎜, ⎟, ⎢ ⎜⎝, n, n2 ⎠, ⎝, ., =, ⎢, n +1 ⎛, n → ∞ ⎢ n 2 ⎛1 + 1 ⎞, 3 ⎜1 +, ⎜, ⎟, n2 ⎠, ⎝, ⎝, ⎣, Lt, , =, , ∴ By ratio test,, (b), , ⎞⎟ ⎤, 3n ⎠ ⎥, ⎥, ⎞, 1, n +1 ⎟ ⎥, 3 ⎠⎦, 1, , 1, <1, 3, , ∑u, , n, , is convergent., , a, ⎡ xn, 2n + 1) ⎤, ⎛ un +1 ⎞, (, Lt ⎜, ×, ⎥, ⎟ = Lt ⎢, a, n →∞, x n −1 ⎥⎦, ⎝ un ⎠ n →∞ ⎣⎢ ( 2n + 3), ⎡ 2a n a 1 + 1 a ⎤, ⎢, 2n . x ⎥ = x, = Lt, a, ⎢, ⎥, n →∞, a a, 3, ⎢⎣ 2 n 1 + 2n, ⎥⎦, , (, (, , By ratio test,, , ∑u, , n, , ), ), , convergence if x < 1 and diverges if x > 1., , 2

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Sequences and Series, , 37, , When x = 1, the test fails; Then, un =, , ⎛u, Lt ⎜ n, n →∞ v, ⎝ n, , 1, , ( 2n + 1), , a, , a, , ⎞, 1, ⎛ n ⎞, ⎟ = nLt, ⎜, ⎟ = nLt, →∞ 2n + 1, →∞, ⎝, ⎠, ⎠, 2+ 1, , (, , ∴ By comparison test,, , ∑u, , and, , n, , n, , ), , ∑v, , n, , a, , =, , ; Taking vn =, , 1, we have,, na, , 1, ≠ 0 and finite ( since a > 0 )., 2a, , have same property, , But p –series test, we have, (i), , ∑v, , n, , convergent when a > 1, , and (ii) divergent when a ≤ 1, , ∴ To sum up, (i) x < 1,, , ∑u, , n, , is convergent ∀a ., , is divergent ∀a ., , (ii), , x > 1,, , (iii), , x = 1, a > 1,, , (iv), (c), , ∑u, n, , ∑u, x = 1, a ≤ 1 , ∑ u, , n, , is convergent, and, , n, , is divergent., , ⎡, 1.2.3....n ( n + 1), 4.7.10.... ( 3n + 3) ⎤, u, ×, Lt n +1 = Lt ⎢, ⎥, n →∞ u, n →∞ 4.7.10.... ( 3n + 3 )( 3n + 6 ), 1.2.3....n, n, ⎣, ⎦, ⎡ ( n + 1) ⎤, 1, = Lt ⎢, = <1 ;, ⎥, n →∞ 3 ( n + 2 ), 9, ⎣⎢, ⎦⎥, 2, , ∴ By ratio test,, (d), , ∑u, , n, , is convergent, , ⎡ (1 + 2n +1 ) (1 + 3n ) ⎤, un +1, ⎥, Lt, = Lt ⎢, ×, n +1, n, n →∞ u, n →∞, ⎢⎣ (1 + 3 ) (1 + 2 ) ⎥⎦, n, , 1, , 2, , ⎡ n +1 ⎛, 1 ⎞ n⎛, 1 ⎞⎤, ⎢ 2 ⎜1 + 2n +1 ⎟ 3 ⎜1 + 3n ⎟ ⎥, ⎝, ⎠× ⎝, ⎠⎥, = Lt ⎢, n →∞, 1, 1, ⎛, ⎞, ⎛, ⎞, ⎢ 3n +1 1 +, 2n ⎜1 + n ⎟ ⎥, ⎜, ⎟, n, +, 1, ⎢⎣, ⎝ 3 ⎠, ⎝ 2 ⎠ ⎥⎦, ∴ By ratio test,, , ∑u, , n, , is convergent., , 1, , 2, , ⎛2⎞, =⎜ ⎟, ⎝3⎠, , 1, , 2, , <1, , 2

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Engineering Mathematics - I, , 38, , (e), , ⎡ 3 ( n + 1)3 + 7 ( n + 1)2 5n9 + 11 ⎤, un +1, = Lt ⎢, × 3, × x⎥, Lt, 9, n →∞ u, n →∞, 3, 7, +, n, +, +, 5, 1, 11, n, (, ), ⎢, ⎥⎦, n, ⎣, , (, , ), , (, , 3, ⎡ 3, 2, 1, 1, ⎢ 3n 1 + n + 7 n 1 + n, = Lt ⎢, 9, n →∞, + 11, 5n 9 1 + 1, ⎢, n, ⎣, , (, , (, , ), , ), , (, , ), , 2, , ⎤, ), (, ⎥, n, 5, ×, × x⎥, 3n (1 + 7, ⎥, 3n ), ⎦, 5n9 1 + 11, , 9, , 3, , 3, , ), , ), ), , (, (, , 3, 2, ⎡ 3⎧, ⎤, 1 + 7 1 + 1 ⎫⎬ 5n9 1 + 11, 3, 1, n, +, ⎨, ⎢, ⎥, 9, n, n, 3n, 5n × x ⎥ = x, ⎩, ⎭×, = Lt ⎢, 9, n →∞, 11 ⎫, ⎧, ⎢, ⎥, 3n3 1 + 7 3, + 9⎬, 5n 9 ⎨ 1 + 1, n, 3, n, ⎢⎣, ⎥⎦, n, 5, ⎩, ⎭, ∴ By ratio test, ∑ un converges when x < 1 and diverges when x > 1., , (, , When x = 1, the test fails,, , (, , 3n3 1 + 7, , Then, , Taking, , ), , ), , (, , ), , 7, 3 1 + 3n, un =, = 6, 5n 1 + 11, 5n9 1 + 11 9, 5n, 5n9, u, 1, 3, vn = 6 , we observe that Lt n = ≠ 0, →∞, n, vn 5, n, , (, , 3n, , ), , (, , ), , ∴ By comparison test and p series test, we conclude that, ∴, , ∑u, , n, , ∑u, , n, , is convergent., , is convergent when x ≤ 1 and divergent when x > 1., , Exercise – 1.2, 1. Test the convergency or divergency of the series whose general term is :, , (a), (b), (c), , (d), (e), , xn, ..............................., n, nx n −1 ..........................., ⎛ 2n − 2 ⎞ n −1, ⎜ n, ⎟ x .............., ⎝ 2 +1 ⎠, ⎛ n2 + 1 ⎞ n, ⎜ 2 ⎟ x ..................., ⎝ n −1 ⎠, , n, nn, , ........................, , [Ans : x < 1cgt , x ≥ 1dgt ], [Ans : x < 1cgt , x ≥ 1dgt ], [Ans : x < 1cgt , x ≥ 1dgt ], [Ans : x < 1cgt , x ≥ 1dgt ], [Ans: cgt.]

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Sequences and Series, , 4n. n, nn, , (f), , .........................., , ( n + 1), ( 3 + 1), , [Ans: dgt.], , n, , 3, , (g), , 39, , n, , ......................., , [Ans: cgt.], , 2. Determine whether the following series are convergent or divergent :, , x, x2, x3, +, +, + .............., 1.2 3.4 5.6, x x 2 x3, 1 + 2 + 2 + 2 + .............., 2 3 4, 1, x, x2, +, +, + ......, 1.2.3 4.5.6 7.8.9, x x 2 x3, xn, 1 + + + + ..... 2, + ...., n +1, 2 5 10, 1.2 2.3 3.4, + 2 + 3 + ..............., x, x, x, , (a), (b), (c), (d), (e), , [Ans : x ≤ 1cgt , x > 1dgt ], [Ans : x ≤ 1cgt , x > 1dgt ], [Ans : x ≤ 1cgt , x > 1dgt ], [Ans : x ≤ 1cgt , x > 1dgt ], [Ans : x > 1cgt , x ≤ 1dgt ], , 1.3.4 Raabe’s Test, , Let, , ∑u, , n, , Then, (i) If k > 1,, Proof :, , ⎧⎪ ⎛ un, ⎞ ⎫⎪, − 1⎟ ⎬ = k, ⎪⎩ ⎝ un +1 ⎠ ⎪⎭, , be series of +ve terms and let Lt ⎨n ⎜, n →∞, , ∑u, , n, , is convergent. (ii) If k < 1,, , Consider the series, , ∑u, , n, , is divergent. (The test fails if k = 1), , 1, , ∑v = ∑ n, n, , p, , ⎡⎛ n + 1 ⎞ p ⎤, ⎡⎛ 1 ⎞ p ⎤, ⎡ vn, ⎤, n⎢, − 1⎥ = n ⎢⎜, ⎟ − 1⎥ = n ⎢⎜1 + ⎟ − 1⎥, ⎢⎣⎝ n ⎠, ⎥⎦, ⎢⎣⎝ n ⎠, ⎥⎦, ⎣ vn +1 ⎦, ⎡⎛, ⎞ ⎤, p p ( p − 1) 1, . 2 + .... ⎟ − 1⎥, = n ⎢⎜ 1 + +, 2, n, ⎢⎣⎝ n, ⎠ ⎥⎦, p ( p − 1) 1, ⎧v, ⎫, . + .........., = p+, Lt n ⎨ n − 1⎬ = p, n →∞, 2, n, ⎩ vn +1 ⎭

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Engineering Mathematics - I, , 40, , ⎧ un, ⎫, − 1⎬ = k > 1, n →∞, ⎩ un +1 ⎭, We choose a number ‘p’ ∋ k > p > 1 ; Comparing the series, , Case (i) In this case, Lt n ⎨, , ∑v, , n, , which is convergent , we get that, , ∑u, , n, , ∑u, , n, , with, , will converge if after some, , fixed number of terms, p, , i.e. If,, , i.e., If, , un, v, ⎛ n +1⎞, > n =⎜, ⎟, un +1 vn +1 ⎝ n ⎠, p ( p − 1) 1, ⎛ u, ⎞, n ⎜ n − 1⎟ > p +, . + .........from (1), ∠, u, 2, n, ⎝ n +1 ⎠, ⎛ u, ⎞, Lt n ⎜ n − 1⎟ > p, n →∞, ⎝ un +1 ⎠, , i.e., If k > p, which is true . Hence, , ∑u, , n, , is convergent .The second case also, , can be proved similarly., , Solved Examples, EXAMPLE 43, , Test for convergence the series, , 1 x3 1.3 x5 1.3.5 x 7, . +, . + ....., x+ . +, 2 3 2.4 5 2.4.6 7, , (JNTU 2006, 2008), , SOLUTION, , Neglecting the first tem ,the series can be taken as ,, , 1 x3 1.3 x5 1.3.5 x 7, . +, . +, . + ....., 2 3 2.4 5 2.4.6 7, 1.3.5....are in A.P. nth term = 1 + ( n − 1) 2 = 2n − 1, 2.4.6...are in A.p. nth term = 2 + ( n − 1) 2 = 2n, 3.5.7.....are in A.P nth term = 3 + ( n − 1) 2 = 2n + 1, , ∴, , un ( nth term of the series) =, , 1.3.5.... ( 2n − 1) x 2 n +1, ., 2.4.6.... ( 2n ) 2n + 1

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Engineering Mathematics - I, , 42, , SOLUTION, , Neglecting the first term,, , un =, , 3.6.9....3n, .x n, 7.10.13....3n + 4, , un +1 =, , 3.6.9....3n ( 3n + 3), .x n +1, 7.10.13.... ( 3n + 4 )( 3n + 7 ), , un +1 3n + 3, u, .x ; Lt n +1 = x, =, n →∞ u, 3n + 7, un, n, , ∑u, , ∴ By ratio test,, , n, , is convergent when x < 1 and divergent when x > 1., , When x = 1 The ratio test fails. Then, , un, 3n + 7 un, 4, ;, =, −1 =, 3n + 3, un +1 3n + 3 un +1, ⎧⎪ ⎛ u, ⎞ ⎫⎪, ⎛ 4n ⎞ 4, Lt ⎨n ⎜ n − 1⎟ ⎬ = Lt ⎜, ⎟ = >1, n →∞, n →∞ 3n + 3, ⎝, ⎠ 3, ⎩⎪ ⎝ un +1 ⎠ ⎭⎪, ∴ By Raabe’s test,, , ∑u, , n, , is convergent .Hence the given series converges if x ≤ 1 and, , diverges if x > 1., EXAMPLE 45, , Examine the convergence of the series, , ∞, , 12.52.92.... ( 4n − 3), , n =1, , 42.82.122.... ( 4n ), , ∑, , 2, , 2, , SOLUTION, , un =, , 12.52.92.... ( 4n − 3), 42.82.122.... ( 4n ), , 2, , 2, , ;, , 2, , 12.52.92.... ( 4n − 3) ( 4n + 1), , un +1 =, , 42.82.122.... ( 4n ) ( 4n + 4 ), 2, , 2, , 2, , ( 4n + 1) = 1 (verify), u, Lt n +1 = Lt, 2, n →∞ u, n →∞, ( 4n + 4 ), n, 2, , ∴ The ratio test fails. Hence by Raabe’s test,, , ∑u, , n, , is convergent. (give proof)

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Sequences and Series, , 43, , EXAMPLE 46, , Find the nature of the series, , ∑, , ( n), , 2, , xn , ( x > 0), , 2n, , (JNTU 2003), , SOLUTION, , un, , ( n), =, , 2, , 2n, , n, , .x ; un +1, , ( n + 1), =, , 2, , 2n + 2, , ( n + 1), un +1, x;, =, un, ( 2n + 1)( 2n + 2 ), , .x n +1, , 2, , (, , ), , 2, , n2 1 + 1, un +1, x, n, Lt, .x =, = Lt, n →∞ u, n →∞, 4, 4n 2 1 + 1, 1+ 2, n, 2n, 2n, , (, , ∴ By ratio test,, , ∑u, , n, , )(, , converges when, , ), , x, < 1 , i. e ; x < 4; and diverges when x >4;, 4, , When x = 4, the test fails., , ( 2n + 1)( 2n + 2 ), un, =, 2, un +1, 4 ( n + 1), un, −2n − 2, −1, ;, −1 =, =, 2, 2 ( n + 1), un +1, 4 ( n + 1), , ∴ By ratio test,, Hence, , ∑u, , ∑u, , n, , ⎡ ⎛ u, ⎞ ⎤ −1, Lt ⎢ n ⎜ n − 1⎟ ⎥ =, <1, n →∞, u, 2, n, +, 1, ⎝, ⎠, ⎣, ⎦, , is divergent, , is convergent when x < 4 and divergent when x > 4, , n, , EXAMPLE 47, , Test for convergence of the series, SOLUTION, , un =, , ∑, , 4.7.... ( 3n + 1) n, x, 1.2.3....n, , 4.7.... ( 3n + 1) n, 4.7.... ( 3n + 1)( 3n + 4 ) n +1, x ; un +1 =, x, 1.2.3....n, 1.2.3....n ( n + 1) ., , ⎡ ( 3n + 4 ) ⎤, un +1, = Lt ⎢, .x ⎥ = 3 x, n →∞ u, n →∞, +, n, 1, (, ), n, ⎣, ⎦, Lt, , (JNTU 1996)

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Engineering Mathematics - I, , 44, , ∴ By ratio test, , ∑u, , n, , converges if 3 x < 1 i.e., x <, , 1, 1, and diverges if x > ;, 3, 3, , 1, , the test fails, 3, ⎤, 1 ⎡u, 1, ⎡ −1 ⎤, ⎡ (n + 1)3 ⎤, x = , n ⎢ n – 1⎥ = n ⎢, – 1⎥ = n ⎢, When, =−, ⎥, 4⎞, 3 ⎣ u n +1 ⎦ ⎣ 3n + 4 ⎦, ⎛, ⎣ 3n + 4 ⎦, ⎜3+ ⎟, n⎠, ⎝, ⎡u, ⎤, 1, Lt n ⎢ n − 1⎥ = − < 1, n →∞, 3, ⎣ un +1 ⎦, If x =, , ∴ By Raabe’s test,, ∴, , ∑u, , n, , ∑u, , n, , is divergent., , is convergent when x <, , 1, 1, and divergent when x ≥, 3, 3, , EXAMPLE 48, , Test for convergence 2 +, SOLUTION, , The nth term, , un =, , 3x 4 x 2 5 x3, +, +, + ........... ( x > 0 ), 2, 3, 4, , ( n + 1) x n−1 ;, n, , un +1 =, , (, (, , ( n + 2) xn ;, ( n + 1), , ), ), , (JNTU 2003), , un +1 n ( n + 2 ), =, .x, 2, un, ( n + 1), , n2 1 + 2, un +1, n .x = x, Lt, = Lt, 2, n →∞ u, n →∞ 2, n, n 1+ 1, n, ∴ By ratio test, ∑ un is convergent if x < 1 and divergent if x > 1, If x = 1, the test fails., , ⎡ ( n + 1)2, ⎤, ⎡, ⎤, ⎡ un, ⎤, 1, Lt n ⎢, − 1⎥ = Lt n ⎢, − 1⎥ = Lt n ⎢, ⎥ = 0 <1, n →∞, n →∞, ⎣ un +1 ⎦ n→∞ ⎢⎣ n ( n + 2 ) ⎥⎦, ⎣ n ( n + 2) ⎦, , Then, , ∴ By Raabe’s test, , ∴, , ∑u, , n, , ∑u, , n, , is divergent, , is convergent when x < 1 and divergent when x ≥ 1, , EXAMPLE 49, , Find the nature of the series, , 3 3.6 3.6.9, +, +, + ......∞, 4 4.7 4.7.10, , (JNTU 2003)

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Sequences and Series, , SOLUTION, , un =, , 45, , 3.6.9.....3n ( 3n + 3), 3.6.9.....3n, ; un +1 =, 4.7.10..... ( 3n + 1)( 3n + 4 ), 4.7.10..... ( 3n + 1), , (, (, , ), ), , 3n 1 + 3, un +1 3n + 3, un +1, 3n = 1, =, ; Lt, = Lt, n, →∞, n, →∞, 3n + 4, un, un, 3n 1 + 4, 3n, Ratio test fails., , ⎡ ⎧u, ⎫⎤, ⎡ ⎛ 3n + 4 ⎞ ⎤, − 1⎟ ⎥, Lt ⎢ n ⎨ n − 1⎬⎥ = Lt ⎢ n ⎜, n →∞, n →∞, ⎣ ⎝ 3n + 3 ⎠ ⎦, ⎣ ⎩ un +1 ⎭⎦, n, n, 1, = Lt, = Lt, = <1, n →∞ 3 ( n + 1), n →∞, 3, 3n 1 + 1, n, ∴ By Raabe’s test ∑ un is divergent., , ∴, , (, , ), , EXAMPLE 50, , If p, q > 0 and the series, , 1+, , 1 p 1.3. p ( p + 1) 1.3.5 p ( p + 1)( p + 2 ), +, +, + ...., 2 q 2.4.q ( q + 1) 2.4.6 q ( q + 1)( q + 2 ), , is convergent , find the relation to be satisfied by p and q., SOLUTION, , un =, , 1.3.5..... ( 2n − 1) p ( p + 1) ..... ( p + n − 1), [neglecting 1st term], 2.4.6.....2n q ( q + 1) ..... ( q + n − 1), , un +1 =, , 1.3.5..... ( 2n − 1)( 2n + 1) p ( p + 1) ..... ( p + n − 1)( p + n ), 2.4.6.....2n ( 2n + 2 ) q ( q + 1) ..... ( q + n − 1)( q + n ), , un +1 ( 2n + 1) ( p + n ), =, ;, un, ( 2n + 2 ) ( q + n ), , (, (, , ), ), , (, (, , ⎡, 1, n 1+ p, ⎢ 2 n 1 + 2n, n, un +1, Lt, ., = Lt ⎢, n →∞ u, n →∞, q, 1, n, ⎢ 2 n 1 + 2n n 1 + n, ⎣, , ∴ ratio test fails., , Let us apply Raabe’s test, , ) ⎤⎥⎥ = 1, ) ⎥⎦

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Engineering Mathematics - I, , 46, , ⎡ ⎧⎪ ( q + n )( 2n + 2 ) ⎫⎪⎤, ⎡ ⎛ u, ⎞⎤, Lt ⎢ n ⎜ n − 1⎟ ⎥ = Lt ⎢ n ⎨, − 1⎬⎥, n →∞, n →∞, +, +, u, p, n, n, 2, 1, (, )(, ), ⎪, ⎣ ⎝ n +1 ⎠ ⎦, ⎭⎪⎦⎥, ⎣⎢ ⎩, ⎡ ⎧, ⎫⎤, ⎢ ⎪ 2q ( n + 1) − p ( 2n + 1) + n ⎪⎥, Lt ⎢ n ⎨, ⎬⎥, n →∞, ⎪⎥, 2+ 1, ⎢ ⎪ n2 1 + p, n, n ⎭⎦, ⎣ ⎩, , (, , (, , )(, , ) (, , ), , ), , ⎡ 2q 1 + 1 − p 2 + 1 + 1 ⎤, n, n, ⎥ = 2q − 2 p + 1, Lt ⎢, n →∞ ⎢, ⎥, 2, 2, ⎣, ⎦, 2q − 2 p + 1, Since ∑ un is convergent, by Raabe’s test,, >1, 2, ⇒ q − p > 1 , is the required relation., 2, , Exercise 1.3, ∞, , 1. Test whether the series, , ∑u, , n, , is convergent or divergent where, , 1, , 22.42.62..... ( 2n − 2 ), un =, .x 2 n, 3.4.5..... ( 2n − 1)( 2n ), 2, , [Ans : x ≤ 1cgt , x > 1dgt ], , 2. Test for the convergence the series, ∞, , ∑, 1, , 4.7.10..... ( 3n + 1) n, x, n, , [Ans : x <, , 1, 1, cgt , x ≥ dgt ], 3, 3, , 3. Test for the convergence the series :, , (i), , 22.42 22.42.52.7 2 22.42.52.7 2.82.102, +, +, + ..., 32.32 32.32.62.62 32.32.62.62.92.92, , (ii), , 3.4, 4.5 2 5.6 3, x+, x +, x + ..... ( x > 0 ), 1.2, 2.3, 3.4, , (iii), , 1.3.5..... ( 2n − 1), xn, ., ∑ 2.4.6.....2n ( 2n + 2 ) ( x > 0 ) [Ans : cgt if x ≤ 1 dgt if , x > 1], , [Ans : divergent], , [Ans : cgt if x ≤ 1 dgt if , x > 1]

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Sequences and Series, , ( 1), 1+, , (iv), , 2, , 2, , 47, , ( 2), x+, , 2, , x2, , 4, , ( 3), +, , 2, , x3, , 6, , + ...... ( x > 0 ), [Ans : cgt if x < 4 and dgt if , x ≥ 4 ], , 1.3.5 Cauchy’s Root Test, , Let, , ∑u, , n, , be a series of +ve terms and let Lt un, , 1, , = l . Then, , n, , n →∞, , ∑u, , n, , is convergent when, , l < 1 and divergent when l > 1, Proof : (i) Lt un, , 1, , = l < 1 ⇒ ∃a +ve number ' λ ' ( l < λ < 1) ∋ un, , n, , n →∞, , 1, , n, , < λ , ∀n > m, , un < λ n , ∀n > m, , (or), , λ < 1, ∑ λ n is a geometric series with common ratio < 1 and, , Since, , therefore convergent., Hence, un is convergent., , ∑, , (ii) Lt un, , 1, , n, , n →∞, , = l >1, , ∴ By the definition of a limit we can find a number r ∋ un, i.e., un > ∀n > r, , 1, , > 1∀n > r, , n, , i.e., after the 1st ‘r ‘ terms , each term is > 1., Lt, un = ∞ ∴, un is divergent., n →∞, , When, , Note :, , Lt ( u, , n →∞, , ∑, , ∑, , 1, n, , n, , ) = 1, the root test can’t decide the nature of ∑ u, , n, , . The fact of, , this statement can be observed by the following two examples., 1., 2., , Consider the series, Consider the series, , ∑, , 1, , n3, , : − Ltun, , 1, , n, , n →∞, , ∑ 1 n , in which Ltu, n →∞, , In both the examples given above,, , Ltu, , n →∞, , 1, n, , n, , 1, n, , n, , 1, , 3, , ⎛ 1 ⎞, = Lt ⎜ 1 ⎟ = 1, n →∞ ⎝ n n ⎠, 1, = Lt 1 = 1, n →∞ n n, , ⎛ 1⎞, = Lt ⎜ 3 ⎟, n →∞ ⎝ n ⎠, , n, , = 1 . But series (1) is convergent, , (p-series test), And series (2) is divergent. Hence when the limit=1, the test fails.

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Engineering Mathematics - I, , 48, , Solved Examples, EXAMPLE 51, Test for convergence the infinite series whose nth terms are:, (i), , 1, n2n, , (ii), , 1, (log n) n, , (iii), , 1, ⎡ 1⎤, ⎢⎣1 + n ⎥⎦, , (JNTU 1996, 1998, 2001), , n2, , SOLUTION, , (i), , (ii), , (iii), , 1, 1, 1, 1, 1, , un n = 2 ; Lt un n = Lt 2 = 0 < 1;, 2n, n →∞, n →∞ n, n, n, By root test ∑ un is convergent., un =, , 1, 1, 1, 1, 1, ; Lt un n = Lt, = 0 < 1;, ; un n =, n, →∞, →∞, n, n, (log n), log n, log n, ∴ By root test , ∑ un is convergent., un =, , un =, , 1, ⎛ 1⎞, ⎜1 + ⎟, ⎝ n⎠, , n2, , ∴ By root test, , ; un, , 1, , n, , ∑u, , n, , =, , 1, ⎛ 1⎞, ⎜1 + ⎟, ⎝ n⎠, , n, , Ltu, , n →∞, , 1, n, , n, , = Lt, , n →∞, , 1, ⎛ 1⎞, ⎜1 + ⎟, ⎝ n⎠, , n, , =, , 1, < 1;, e, , is convergent., , EXAMPLE 52, , Find whether the following series are convergent or divergent., ∞, , 1, (i) ∑ n, n =1 3 − 1, , (iii), , SOLUTION, , (i), , un, , 1, , n, , ⎛ 1 ⎞, =⎜ n ⎟, ⎝ 3 −1 ⎠, , 1, , n, , ⎛, ⎞, ⎜, ⎟, 1, ⎟, =⎜, ⎜ 3n ⎛1 − 1 ⎞ ⎟, ⎜ ⎜ 3n ⎟ ⎟, ⎠⎠, ⎝ ⎝, , ⎡⎣( n + 1) x ⎤⎦, ∑, n n +1, n =1, ∞, , 1 1 1, (ii) 1 + 2 + 3 + 4 + ....., 2 3 4, , 1, , n, , n

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Sequences and Series, , Lt un, , 1, , n, , n →∞, , 49, , ⎛, ⎜, 1, = Lt ⎜, n →∞, 1, ⎜ n⎛, ⎜ 3 ⎜1 − 3n, ⎝ ⎝, , ⎞, ⎟, ⎟, ⎞⎟, ⎟⎟, ⎠⎠, , 1, , n, , 1, 1, ⎛ 1 ⎞, un = n ; Lt un n = Lt ⎜ n ⎟, →∞, →∞, n, n, n, ⎝n ⎠, , (ii), , (iii), , ⎡( n + 1) x ⎤⎦, un = ⎣, n n +1, Lt un, , 1, , n, , n →∞, , 1, < 1 ; By root test,, 3, , ∑u, , = 0 < 1 ; By root test,, , ∑u, , =, , 1, , n, , n, , n, , is convergent., , is convergent., , n, , ⎡ {( n + 1) x}n ⎤, ⎥, = Lt ⎢, n →∞ ⎢, n n +1, ⎥, ⎣, ⎦, , 1, , n, , 1, , ⎡ ⎧ ( n + 1) x ⎫n 1 ⎤ n, ⎛ n +1⎞ 1, Lt ⎢ ⎨, ⎬ . ⎥ = nLt, ⎜, ⎟ x. 1, →∞, n →∞, ⎝ n ⎠ n n, ⎢⎣ ⎩ n ⎭ n ⎥⎦, ⎛, ⎞, 1, 1, ⎛ 1⎞ 1, Lt ⎜ 1 + ⎟ x. 1 = Lt x. 1 = x, since Lt x. 1 = 1⎟, ⎜, n →∞, n →∞, n →∞, ⎝ n⎠ n n, n n, n n, ⎝, ⎠, ∴ ∑ un is convergent if x < 1 and divergent if x > 1 and when x = 1 the, test fails., Then un, , ( n + 1), =, n, , n, , vn =, , ; Take, , n +1, , 1, n, , ( n + 1) = ⎛1 + 1 ⎞ ;, un ( n + 1), =, .n =, ⎜, ⎟, n +1, vn, n, nn, ⎝ n⎠, ∴ By comparison test, ∑ un is divergent., n, , ∑v, Hence ∑ u, (, , n, , If un =, , diverges by p –series test ), , n, , is convergent if x < 1 and divergent x ≥ 1, , 2, , ( n + 1), , n2, , un, = e >1, n →∞ v, n, Lt, , n, , EXAMPLE 53, , nn, , n, , , show that, , ∑u, , n, , is convergent.

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Engineering Mathematics - I, , 50, , Lt un, , 1, , n, , n →∞, , ⎡, n2, n, = Lt ⎢, n2, n →∞ ⎢, +, 1, n, (, ), ⎣, , ⎤, ⎥, ⎥, ⎦, , 1, , n, , nn, , ; = Lt =, , ( n + 1), , n →∞, , n, , ⎛ n ⎞, = Lt ⎜, ⎟, n →∞ n + 1, ⎝, ⎠, , n, , n, , ⎛, ⎞, ⎜ 1 ⎟ 1, = Lt ⎜, ⎟ = <1; ∴, n →∞, e, ⎜ 1+ 1 ⎟, ⎝ n⎠, , ∑u, , n, , converges by root test ., , EXAMPLE 54, 2, , 3, , 1 ⎛2⎞ ⎛3⎞, Establish the convergence of the series, + ⎜ ⎟ + ⎜ ⎟ + ........, 3 ⎝5⎠ ⎝7⎠, SOLUTION, n, , ⎛ n ⎞, un = ⎜, ⎟ .........(verify);, ⎝ 2n + 1 ⎠, By root test, ∑ un is convergent., , Lt un, , n →∞, , 1, , n, , ⎛ n ⎞ 1, = Lt ⎜, ⎟ = <1, n →∞ 2 n + 1, ⎝, ⎠ 2, , EXAMPLE 55, ∞, , Test for the convergence of, , ∑, n =1, , n n, .x, n +1, , SOLUTION, 1, , 1, , ⎛, ⎞2, ⎛, ⎞2, ⎜, ⎜ 1 ⎟ n, 1, 1 ⎟, .x ; Lt un n = Lt ⎜, un = ⎜, ⎟ .x = x, ⎟, n →∞, n →∞, 1, 1, ⎜ 1+ ⎟, ⎜ 1+ ⎟, ⎝ n⎠, ⎝ n⎠, ∴ By root test, ∑ un is convergent if x < 1 and divergent if x > 1 ., When x = 1 : un =, , 1, n, , taking vn = 0 and applying comparison test , it can be, n, n +1, , seen that is divergent, un is convergent if x < 1 and divergent if x ≥ 1 ., , ∑, , EXAMPLE 56, , ∑(, ∞, , Show that, , n =1, , 1, , ), , n, , n n − 1 converges.

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Sequences and Series, , SOLUTION, , (, , un = n, , Lt un, , 1, , 1, , n, , n →∞, , ∑u, , ∴, , n, , 51, , ), , −1, , n, , (, , ), , 1, , (, , = Lt n n − 1 = 1 − 1 = 0 < 1 since Lt n, n →∞, , 1, , n, , n →∞, , ), , =1 ;, , is convergent by root test., , n, , EXAMPLE 57, n, , Examine the convergence of the series whose n, , th, , ⎛n+2⎞ n, term is ⎜, ⎟ .x, ⎝ n+3⎠, , SOLUTION, n, , 1, ⎛n+2⎞ n, ⎛n+2⎞, un = ⎜, un n = Lt ⎜, ⎟x = x, ⎟ .x ; nLt, n, →∞, →∞, ⎝ n+3⎠, ⎝ n+3⎠, ∴ By root test, ∑ un converges when x < 1 and diverges when x > 1 ., n, , ⎛ 2⎞, n, ⎜1 + ⎟, n⎠, ⎛n+2⎞, un = Lt ⎝, When x = 1 : un = ⎜, ⎟ ; nLt, n, →∞, →∞, n, ⎝ n+3⎠, ⎛ 3⎞, 1, +, ⎜, ⎟, ⎝ n⎠, e2 1, = ≠ 0 and the terms are all +ve ., =, e3 e, ∴ ∑ un is divergent . Hence ∑ un is convergent if x < 1 and divergent if x ≥ 1 ., EXAMPLE 58, , Show that the series,, −1, , −2, , −3, , ⎡ 22 2 ⎤, ⎡ 33 3 ⎤, ⎡ 44 4 ⎤, −, +, −, +, ⎢ 12 1 ⎥, ⎢ 23 2 ⎥, ⎢ 34 − 3 ⎥ + ...... is convergent, ⎣, ⎦, ⎣, ⎦, ⎣, ⎦, −n, , ⎡ ( n + 1)n +1 n + 1 ⎤, −, un = ⎢, ⎥ ;=, n +1, n ⎥⎦, ⎢⎣ n, ⎛ 1⎞, ⎜1 + ⎟, ⎝ n⎠, , −n, , −n, , ⎛ n +1⎞, ⎜, ⎟, ⎝ n ⎠, , −n, , ⎡⎛ n + 1 ⎞ n ⎤, ⎢⎜, ⎟ − 1⎥, ⎢⎣⎝ n ⎠, ⎥⎦, , (JNTU 2002), , −n, , −1, n, ⎡⎛ 1 ⎞ n ⎤, ⎤, 1, ⎛ 1 ⎞ ⎡⎛ 1 ⎞, n, +, −, =, +, +, ;, u, 1, 1, 1, 1, ⎢⎜, ⎥, n, ⎜, ⎟ ⎢⎜, ⎟ − 1⎥, ⎟, ⎝ n ⎠ ⎢⎣⎝ n ⎠, ⎥⎦, ⎢⎣⎝ n ⎠, ⎥⎦, , −1

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Engineering Mathematics - I, , 52, , =, , 1, , 1, , ⎛ 1 ⎞ ⎪⎧⎛ 1 ⎞ n ⎪⎫, ⎜1 + ⎟ 1 +, − 1⎬, ⎝ n ⎠ ⎨⎪⎜⎝ n ⎟⎠, ⎩, ⎭⎪, 1, 1 1, 1, ∴ Lt un n = ., =, <1, n →∞, 1 e −1 e −1, ∴ By root test, ∑ un is convergent., EXAMPLE 59, ∞, , Test, , ∑u, m =1, , m, , for convergence when um =, , e− m, , (, , 1+ 2, , m, , ), , − m2, , SOLUTION, , ( ), , Lt um, , m →∞, , 1, , m, , (, , ⎡, 2, ⎢ 1+ m, = Lt ⎢, m →∞, em, ⎢, ⎣, , Hence Cauchy’s root tells us that, , ), , m2, , ∑u, , 1, , ⎤, ⎥, ⎥, ⎥, ⎦, , m, m, , 1⎛, 2⎞, e2, 1, +, =, = e >1, ⎜, ⎟, m →∞ e, e, ⎝ m⎠, , ; Lt, , is divergent., , m, , EXAMPLE 60, , Test the convergence of the series, , n, , ∑e, , n2, , ., , SOLUTION, 1, , Lt un, , 1, , n →∞, , n, , n n, = Lt n = 0 < 1, n →∞ e, , ∴ By root test,, , ∑u, , n, , is convergent., , EXAMPLE 61, , ( n + 1) .x + ......, x > 0, 2, 32, Test the convergence of the series, 2 x + 3 x 2 + ..., 1, 2, n n +1, n, , n, , SOLUTION, , Lt un, , n →∞, , 1, , n, , ⎡ ( n + 1)n .x n ⎤, = Lt ⎢, ⎥, n +1, n →∞, ⎢⎣ n, ⎥⎦, , 1, , n, , ⎡⎛ n + 1 ⎞ 1 ⎤, = Lt ⎢⎜, ⎟ . 1 n .x ⎥, n →∞, n, ⎝, ⎠ n, ⎣, ⎦

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Sequences and Series, , 53, , 1, ⎡⎛ 1 ⎞ 1 ⎤, = Lt ⎢⎜1 + ⎟ . 1 .x ⎥ = 1.1.x = x ⎡since Lt n n = 1⎤, ⎢, ⎥⎦, n →∞, n →∞, ⎣, ⎣⎝ n ⎠ n n ⎦, ∴ By root test, ∑ un converges if x < 1 and diverges when x > 1 ., When x = 1 , the test fails., n, , 1, ⎛ 1⎞ 1, un = ⎜ 1 + ⎟ . ; Take vn =, n, ⎝ n⎠ n, , Then, , n, , un, ⎛ 1⎞, = Lt ⎜1 + ⎟ = e ≠ 0, n →∞ v, n →∞, ⎝ n⎠, n, Lt, , ∴ By comparison test and p-series test,, Hence, , ∑u, , ∑u, , n, , is divergent., , is convergent when x < 1 and divergent when x ≥ 1 ., , n, , Exercise 1.4, 1. Test for convergence the infinite series whose nth terms are:, (a), (b), , 1, 2 −1, 1, , ............................., , n, , ( log ), , . ( n ≠ 1) ......................, , 2n, , [Ans : convergent], [Ans : convergent], , n, , (c), (d), (e), (f), , (g), , (h), , ⎛ 3n + 1 ⎞, .x ⎟ ................................., ⎜, ⎝ 4n + 3 ⎠, xn, ............................................, n, n, ............................................., nn, 3n.∠n, ........................................, n3, , ( 2n, , (, , 2, , − 1), , ( 2n ), n, , 1, , n, , ), , 4, 4, cgt , x ≥ dgt ], 3, 3, , [Ans : cgt for all x ≥ 0 ], [Ans : convergent], [Ans : convergent], , n, , 2n, , −1, , [Ans : x <, , ..................................., , [Ans : convergent], , ..................................., , [Ans : convergent], , 2n