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Unformatted text preview: Homework answers, week of Sept. 9 1. Proof #1. Since { x n } diverges to infinity, then given B > 0, there exists N > 0 such that x n > B for all n > N . The hypotheses imply that y n ≥ x n > B for all n > N . Hence { y n } → ∞ . Proof #2. Suppose that { y n } were bounded. Then { y n } has a least upper bound, say B . Therefore, x n ≤ y n ≤ B for all n , which implies that { x n } is bounded, a contradiction. Hence { y n } does not have an upper bound. 2. The problem doesn’t ask for a rigorous proof, just a determination of whether the given sequence is Cauchy. I have indicated a sample justification for each answer. (a) Problem 2, p. 81. The sequence x n = ∑ n i = 1 ( 1 / i ! ) is Cauchy. One pos sible rationale:  x n + k x n  < k ∑ j = 1 1 ( n + 1 ) j = 1 n 1 1 ( n + 1 ) k < 1 n . Given ε , we may choose n so that  x n + k x n  < ε for all k ∈ N . Hence the sequence is Cauchy....
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 Spring '07
 thieme
 Order theory, Limit of a sequence, Cauchy sequence, Cauchy, Howard Staunton

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