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RA-L 2.4, Mr. Anil Singh, Assistant Professor, Department of Mathematics, Govt. MAM College,Jammu, , Mr. Anil Singh Assistant Professor, , RA-L 2.4

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The following examples illustrate how the definition is applied to, prove that a sequence has a particular limit. In each case, a, positive  is given and we are required to find a K, depending on ,, as required by the definition., Example 1., lim (1/n) = 0., If  > 0 is given, then 1/ > 0. By the Archimedean Property ,, there is a natural numberK() such that 1/K < . Then, if, n ≥ K, we have 1/n ≤ 1/K < ., Consequently, if n ≥ K, then, 1, 1, − 0 = < ., n, n, Therefore, we can assert that the sequence (1/n) converges to 0., , Mr. Anil Singh Assistant Professor, , RA-L 2.4

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Example 2., lim n21+1 = 0., Let  > 0. To find K, we first note that if n ∈ N, then, n2, , 1, 1, 1, < 2 ≤ ., +1, n, n, , Now choose K > 0 such that 1/K < , as in Example 1. Then, n ≥ K implies that 1/n < , and therefore, n2, , 1, 1, 1, −0 = 2, < < ., +1, n +1, n, , Hence, we have shown that the limit of the sequence is zero., , Mr. Anil Singh Assistant Professor, , RA-L 2.4

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Example 3., If 0 < b < 1, then lim (bn ) = 0., We will use elementary properties of the natural logarithm, function. If  > 0 is given, we see that, bn <  ⇐⇒ n ln b < ln  ⇐⇒ n > ln / ln b., (The last inequality is reversed because ln b < 0). Thus if we, choose K to be a number such K > ln / ln b, then we have, 0 < bn <  for all n ≥ K. Thus we have that lim (bn ) = 0., , Mr. Anil Singh Assistant Professor, , RA-L 2.4

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Exercises: Show that, 1, 2, , lim 3n+2, n+1 = 3., √, √
, lim n + 1 − n = 0., , Mr. Anil Singh Assistant Professor, , RA-L 2.4

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In order to show that a sequence X = (xn ) does not converge to, the number x, it is enough to produce one number 0 > 0 such, that no matter what natural number K is chosen, one can find a, particular nK satisfying nK ≥ K such that |xnK − x| ≥ 0 ., Example 4., The sequence (xn ) = (0, 2, 0, 2 · · · , 0, 2, · · · ) does not converge to, the number 0., Suppose on the contrary that 0 is the limit of the given sequence., Then for 0 <  < 2, there exists K > 0 such that for n ≥ K,, |xn − 0| = |xn | < . In prticular |xK | <  and |xK+1 | < . Now, either xK = 2 or xK+1 = 2, in both the case we get 2 < , which, is a contradiction to the choice of ., , Mr. Anil Singh Assistant Professor, , RA-L 2.4

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It is important to realize that the convergence (or divergence) of a, sequence X = (xn ) depends only on the ultimate behavior of the, terms. By this we mean that if, for any natural number m, we, drop the first m terms of the sequence, then the resulting sequence, Xm converges if and only if the original sequence converges, and in, this case, the limits are the same. We will state this formally after, we introduce the idea of a ’tail’ of a sequence., Definition 5., If X = (x1 , x2 , · · · ) is a sequence of real numbers and if m is a, given natural number, then the m−tail of X is the sequence, Xm := (xm+n : n ∈ N) ., , Mr. Anil Singh Assistant Professor, , RA-L 2.4

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Theorem 6., Let X = (x1 , x2 , · · · ) is a sequence of real numbers and let m be a, given natural number. Then the m−tail Xm := (xm+n : n ∈ N) of, X is convergent if and only if the sequence X converges. In this, case lim Xm = lim X., Proof: We note that for any p ∈ N, the pth term of the tail Xm is, the (p + m)th term of the sequence X. Similarly, if q > m, the q th, term of X is the (q − m)th term of Xm ., Assume X converges to x. Then given any  > 0, if the terms of, X for n ≥ K() satisfy |xn − x| < , then the terms of Xm for, k ≥ K() − m satisfy |xk − x| < . Thus we can take, Km () = K() − m, so that Xm also converges to x., Conversely, if the terms of Xm for k ≥ Km () satisfy |xk − x| < ,, then the terms of X for n ≥ Km () + m satisfy |xn − x| < . Thus, we can take K() = Km () + m., Mr. Anil Singh Assistant Professor, , RA-L 2.4