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The third difficulty is proving the sequence actually

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Unformatted text preview: The third difficulty is proving the 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 | 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 | ....
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The third difficulty is proving the sequence actually...

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