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# HW4 - R satisfying | a n | 6 b n-b n 1 for all n ∈ N...

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ANSWERS TO SELECTED EXERCISES: HW4 Exercise 7.3.3a Prove that if a k converges and b k converges absolutely, then a k b k converges. Proof : Since a k converges, we must have lim n →∞ a n = 0, thus in particular a n is bounded, say | a n | < M for all n N , for some M > 0. We will prove convergence of the series a k b k using the Cauchy Criterion (Theorem 2.7.3); let p 6 q be some positive integers. Then q X k = p a k b k 6 q X k = p | a k | | b k | < M q X k = p | b k | . Since b k converges absolutely, then for all ε > 0 there exists some positive integer n 0 such that q X k = p | b k | < ε/M for all q > p > n 0 by the Cauchy Criterion. By the above inequalities, we will have q X k = p a k b k < ε for all q > p > n 0 , hence the same theorem yields the convergence of the series a k b k . Exercise 7.3.4 Suppose that the sequence { b n } is monotone decreasing with lim b n = 0. If { a n } is a sequence in R satisfying

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Unformatted text preview: R satisfying | a n | 6 b n-b n +1 for all n ∈ N , prove that ∑ a k converges absolutely. Proof : We will use again the Cauchy Criterion; let p 6 q be arbitrary positive integers. Then q X k = p | a k | 6 b p-b q +1 , by hypothesis. Since { b n } is monotone decreasing and convergent, hence Cauchy, for arbitrary ε > 0 there exists some n ∈ N such that for all q > p > n we have | b p-b q +1 | = b p-b q +1 < ε. Thus, for all q > p > n , we have q X k = p | a k | < ε 1 thus ∑ a k converges absolutely by the Cauchy Criterion. Remark Note that the hypothesis lim b n = 0 was not used! We only used the fact that b n is convergent. 2...
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HW4 - R satisfying | a n | 6 b n-b n 1 for all n ∈ N...

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