This preview shows pages 1–2. Sign up to view the full content.
This preview has intentionally blurred sections. Sign up to view the full version.
View Full Document
Unformatted text preview: Modern Analysis, Homework 6 Possible solutions 1. [15] (5 pts) Rudin, Ch 3, problem 4. Solution: First I claim that s 2 m = 1 2 1 2 m 1 and that s 2 m +1 = 1 1 2 m . Lets prove this by induction. Let P ( n ) be the statement: s n = 1 2 1 2 n/ 2 if n is even, or s n = 1 1 2 n 1 2 if n is odd. Clearly P (0) holds. Assume P (0) , ,P ( k ) for some k N . First assume that k is even. Then by induction we have s k = 1 2 1 2 k/ 2 . By definition, s k +1 = 1 2 + s k since k + 1 is odd. Hence s k +1 = 1 2 + 1 2 1 2 k/ 2 = 1 1 2 ( k +1) 1 2 . Therefore P ( k +1) holds. A similar argument shows that P ( k + 1) holds in the case that k is odd as well. At this point it is clear that lim sup n s n = 1 and lim inf n = 1 2 . In order to argue this rigorously, lets use Rudins characterization of lim sup, Theorem 3.17. To prove the statement about lim sup s n , there are two things to show: 1. 1 is a subsequential limit of s n 2. If x > 1, then there is an integer N such that n > N implies that s n < x . Item 1 is clear: s 2 n +1 1. Consider item 2: Let x > 1. Then for any n N , we either have s n = 1 1 2 m 1 2 < 1 (if n is odd) or s n = 1 2 1 2 m/ 2 < 1 (if n is even). In any case, we always have s n < 1. Therefore lim sup n s n = 1. To prove the statement about lim inf s n there are two things to prove: 1. 1 2 is a subsequential limit of s n 2. If x < 1 2 then there is an integer N such that n > N implies that s n > x Item 1 is clear: s 2 n 1 2 . Consider item 2: Let x < 1 2 . Choose N N such that 1 / 2 N < 1 2 x . Suppose that n > N . If n = 2 k , then s 2 k = 1 2 1 2 k > 1 2 + x 1 2 = x . Likewise, if n = 2 k + 1, then s 2 k +1 = 1 1 2 m > 1 2 > x . 2. [23] (5 pts) Rudin, Ch 3, problem 5. Solution: Suppose that it is not the case that lim sup n a n = and lim sup n b n = (or that lim sup n a n = and lim sup n b n = + ). Then the expression lim sup n a n + lim sup n b n makes sense as an exended real. Consider the case that lim sup n ( a n + b n ) = . By definition, for any extended real x R { ,} , we have x . It follows that lim sup n ( a n + b n ) lim sup n a n + lim sup n b n . Now consider the case that = lim sup n ( a n + b n ) R . Find a subsequence ( a n k + b n k ) which coverges to . This gives subsequences a n k , b n k of a n , b n , respectfully. These subse quences need not converge, but they either approach or have convergent subsequences,...
View
Full
Document
This note was uploaded on 10/18/2011 for the course MATH S4061X taught by Professor Peters during the Spring '11 term at Columbia.
 Spring '11
 Peters

Click to edit the document details