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s11 - Math 5615H Introduction to Analysis I Homework#11...

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Math 5615H: Introduction to Analysis I. Fall 2011 Homework #11. Problems and Solutions. #1. (12 points). Find S := n =1 n 2 n . Solution. Substituting n = m + 1, we get 2 S = n =1 n 2 n - 1 = m =0 m + 1 2 m = S + m =0 ( 1 2 ) m = S + 2 , so that S = 2. #2. (10 points). Give an example of a convergent series a n , such that n =0 c n diverges, where c n := n k =0 a k a n - k . Solution. See Example 3.49 in the textbook. #3. (14 points). Investigate the behavior (absolute convergence, nonabsolute con- vergence, or divergence) of the series ( a ) n =1 ( - 1) n ln n n ; ( b ) n =1 ln n n 2 . Solution. First we show that for an arbitrary α = const > 0, we have ln n n α 0 as n → ∞ . (1) Indeed, denote x n := ln n . For each natural n 3 > e , there is a natural m such that m x n < m + 1, and m → ∞ as n → ∞ . Therefore, for n 3, 0 < ln n n α = x n e α x n < m + 1 ( e α ) m 0 as n → ∞ by Theorem 3.20(d) with 1 + p = e α > 1. This proves the convergence in (1). (a). Note that a n := ln n n > 1 n for n 3 .

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