HW1-4 - sums. Let P n = { , 1 /n,. .., ( n-1) /n, 1 } . To...

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ANSWERS TO SELECTED EXERCISES: HW1-4 Exercise 2.5.10 Suppose { a n } , { b n } are sequences of nonnegative real numbers with lim n →∞ b n = b 6 = 0 and lim n →∞ a n = a . Prove that lim n →∞ a n b n = ab . Proof : By hypothesis, both { a n } and { b n } are bounded, hence so is { a n b n } . This shows first that lim n →∞ a n b n < . Let { n k } be a strictly increasing sequence of positive integers, such that lim k →∞ a n k b n k = lim n →∞ a n b n . Since lim k →∞ b n k = b 6 = 0, then lim k →∞ a n k = lim k →∞ a n k b n k b n k = 1 b lim n →∞ a n b n . Now, the limit of any convergent subsequence of { a n } is less than or equal to a , therefore the following must hold lim n →∞ a n b n 6 ab. On the other hand, let { a n l } be a subsequence of { a n } converging to a . Then, { a n l b n l } is a convergent subsequence of { a n b n } , converging of course to ab . But this clearly shows that lim n →∞ a n b n > ab, thus finally we get lim n →∞ a n b n = ab. Execise 6.2.6 Let f be continuous on [0 , 1]. Prove that lim n →∞ 1 n n X k =1 f ± k n ² = Z 1 0 f ( x ) dx. Proof : We know that the integral on the right-hand side exists, since f is continuous. Now, we observe that the left-hand side is the limit of Riemann
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Unformatted text preview: sums. Let P n = { , 1 /n,. .., ( n-1) /n, 1 } . To form a Riemann sum with respect to this partition, we choose a point in each interval [( i-1) /n,i/n ], namely i/n . Then we observe that the left-hand side is equal to lim n S ( P n ,f ) . 1 Since lim n kP n k = 0, the above must tend to R 1 f ( x ) dx , as desired. Exercise 7.2.8 Suppose that a k converges. Prove that lim n 1 n n X k =1 ka k = 0 . Proof : Using the Abel partial summation formula (Theorem 7.2.1) for p = 1, q = n and b k = k , we get n X k =1 ka k = nA n-n-1 X k =1 A k . Therefore, lim n 1 n n X k =1 ka k = lim n ( A n-1 n n-1 X k =1 A k ) = A-A = 0 , by exercise 2.2.14. 2...
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HW1-4 - sums. Let P n = { , 1 /n,. .., ( n-1) /n, 1 } . To...

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