The second and third difficulty are frequently

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sequence actually converges to the assumed limit. The second and third difficulty are frequently interchangeable. We just overcame the first difficulty; if the sequence converges to anything it has to be to a . Back to the proof. The sequence { a n } is a Cauchy sequence, thus is bounded. (recall that Cauchy sequences are bounded!). In other words, there is some real number C > 0 such that k a n k 1 C for n = 1 , 2 , 3 . . . . For every K N we have K X k =1 | a nk | ≤ X k =1 | a nk | = k a n k 1 C for n = 1 , 2 , 3 , . . . . Since (it being a finite sum!) lim n →∞ K X k =1 | a nk | = K X k =1 | lim n →∞ a nk | = K X k =1 | a k |
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we see that K X k =1 | a k | ≤ C for all K , hence also k a k 1 = X k =1 | a k | ≤ C, proving a 1 . Finally, we need to show that k a n - a k 1 0 as n → ∞ . The problem her is to control well the tails of the sequences. Let ² > 0 be given. Because a 1 , there is K such that X k = K +1 | a k | < ² 2 . By the Cauchy condition, there is N such that k a n - a m k 1 < ²/ 2 whenever n, m N . Thus, for every p 1, n, m N , K + p X k = K +1 | a nk - a mk | ≤ k a n - a m k 1 < ² 2 . Keeping p fixed for a moment we let m → ∞ (the sum being finite, the limit can be interchanged with the sum) to conclude that K + p X k = K +1 | a nk - a k | < ² 2 for all p N , n N , n N . This being true for all p 1 it is also true for the limit as p → ∞ ; in other words, X k = K +1 | a nk - a k | < ² 2 if n N . Now (again because the sum is finite) lim n →∞ K X k =1 | a nk - a k | = 0 , thus there is N 1 such that K X k =1 | a nk - a k | < ² 2 if n N 1 . Let n max( N, N 1 ). Then k a n - a k 1 = X k =1 | a nk - a k | = K X k =1 | a nk - a k | + X k = K +1 | a nk - a k | < ² 2 + ² 2 = e proving that { a n } converges to a in 1 . NOTES: 1. If you got only 5 points for this exercise it is because I assumed that everybody would have no difficulty proving everything but the completeness, thus I emphasized that part of the problem. A 5 means that, in my opinion, your proof of completeness has profound flaws. But see also Note 3. 2. Several of you used (essentially) the following result without justification: If lim n →∞ a nk = a k , then (1) X k =1 | a k | = lim n →∞ X k =1 | a nk | . 2
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Exchanging infinite processes is a delicate operation; it has to be done with care. For example, suppose we take a nk = 1 n 2 k if k 6 = n , k k=n For a fixed k , once n > k the sequence is the same as 1 / (2 k n ) 0 as n → ∞ , thus a k = lim n →∞ a nk = 0 for all k .
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