However it should be clear that x k 1 a nk a kk k for

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However, it should be clear that X k =1 a nk > a kk = k for all n . So we have here a situation where each series k =1 a nk converges (absolutely since all terms are positive), the series coming from the limit sequence { a k } converges even more so since all terms are 0, but (1) is quite false. 3. Several of you got 5 for this exercise for using a correct proof incorrectly. It gave me the impression that you were not quite sure of what you were doing. Before using that proof, a lot of things have to be proved. Here is a sketch of how it works. Let X be a normed vector space. Because it is a vector space, n k =1 a k makes sense for any finite number a 1 , . . . , a k X . If a n X for n = 1 , 2 , 3 , . . . , one can talk now of the series n =1 a n and say that it converges iff and only if the sequence { n k =1 a k } n =1 converges, and write a = n =1 a n iff lim n →∞ n k =1 a k = a ; i.e., if lim n →∞ k n k =1 a k - a k = 0. We say the series converges absolutely iff the series of nonnegative real terms n =1 k a n k converges. Then one has: Theorem 1 If X is a complete normed space; i.e., a Banach space, absolute convergence implies convergence. The proof is similar (or identical) to the proof in the complex case, but it requires a proof. Another result needed for this other proof of completeness, a result that cannot be used so freely, that requires a proof, is one that is valid in all metric spaces, namely: Lemma 2 Let { p n } be a Cauchy sequence in a metric space M . Then there is a subsequence { p n k } such that d ( p n k - 1 , p n k ) < 2 - k . The proof is easy, but at our level it is needed. The main result involved is the converse of Theorem 1, namely: Theorem 3 If X is a normed vector space in which every absolutely convergent sequence converges, then X is a Banach space. People who used this approach provided what amounts to a moderately incorrect proof of this result. All in all, I felt there were too many gaps, and several errors, in the proofs i saw, to warrant more points than I gave. 3. Let { f n } be a sequence of continuously differentiable functions on [0 , 1] with f n (0) = f 0 n (0) and | f 0 n ( x ) | ≤ 1 for all x [0 , 1] and n N . Show that if lim n →∞ f n ( x ) = f ( x ) for all x [0 , 1], then f is continuous on [0 , 1]. Must the sequence converge? Must there be a convergent subsequence?
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