s11 - Math 5615H: Introduction to Analysis I. Fall 2011...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
This is the end of the preview. Sign up to access the rest of the document.

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....
View Full Document

This document was uploaded on 02/15/2012.

Page1 / 2

s11 - Math 5615H: Introduction to Analysis I. Fall 2011...

This preview shows document pages 1 - 2. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online