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Unformatted text preview: 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 = const > 0, we have ln n n 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, < ln n n = x n e x n < m + 1 ( e ) m as n by Theorem 3.20(d) with 1 +by Theorem 3....
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This document was uploaded on 02/15/2012.
 Fall '09
 Math

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