This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: (1) (a) Give a sequence { a n } such that ∑ n ≥ a n converges, but Q n ≥ (1+ a n ) does not. (b) Give a sequence { a n } such that Q n ≥ (1+ a n ) converges, but ∑ n ≥ a n does not. (c) Show that if ∑ n ≥  a n  2 converges, then ∑ n ≥ a n converges if and only if Q n ≥ (1 + a n ) converges. Solution: (a) Let b n be any decreasing sequence, converging to 0, i.e. b ≥ b 1 ≥ b 2 ≥ ··· ≥ 0. Then ∑ n ≥ ( 1) n b n converges. But Q n ≥ (1 + ( 1) n b n ) does not necessarily converge, for example take b n = 1 / p 1 + b n/ 2 c , then Q 2 m n =0 (1 + ( 1) n b n ) = Q m n =0 (1 1 / ( n + 1)) = 1 m +1 , which converges to 0 (so the infinite product diverges). (b) Take a 2 n = 1 / √ n + 1 and a 2 n +1 = 1 / (1+1 / √ n + 1) 1, so Q 2 m 1 n =0 (1+ a n ) = 1 and Q 2 m n =0 (1+ a n ) = 1+ a 2 m = 1+1 / √ m + 1 which converges to 1. However the infinite series ∑ n ≥ a n diverges as a 2 n + a 2 n +1 = 1 √ n + 1 + 1 1 + 1 √ n +1 1 = 1 √ n + 1 + √ n + 1 1 + √ n + 1 1 = 1 √ n + 1 1 1 + √ n + 1 = 1 √ n + 1(1 + √ n + 1) = 1 n + 1 + √ n + 1 so ∑ n ≥ ( a 2 n + a 2 n +1 ) diverges. (c) First note that if ∑ n ≥  a n  2 converges, then a n → 0 as n → ∞ , so after some finite time we will not have any a n = 1 and thus we can assume without loss of generality that this never occurs. We can even assume that  a n  < 1 / 2 for all n . In particular we can take a branch of log( z + 1) = R z 1 / (1 + t ) dt on D (0 , 1) and observe that log(1 +...
View
Full
Document
This document was uploaded on 12/08/2010.
 Spring '09

Click to edit the document details