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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 . Let’s 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, let’s use Rudin’s 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 α = lim sup n →∞ ( a n + b n ) ∈ R . Find a subsequence ( a n k + b n k ) which coverges to α . 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,...
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 Spring '11
 Peters
 lim, lim sup, subsequence, lim sup bn

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