If I have a limit of an infinite sum, can I bring the limit inside the infinite sum? If so, why?
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Consider the limit $$\lim\limits_{x\to-1}\sum_{n=1}^{\infty}\frac{(-1)^{n+1}x^n}{n}$$ evaluating limit from inside gives us $\sum_{n=1}^{\infty}\frac{(-1)^{2n+1}}{n}=\ln 2$. But, evaluating the limit from outside gives us $\lim\limits_{x\to-1}\ln(1+ x)\to-\infty$. So we cannot interchange limits. For a more broad discussion, see here
vidyarthi
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What is the reason why the limit cannot be brought inside the sum, other than that two different answers are obtained with two different methods of evaluation? Thanks! – Math12345 Apr 22 '17 at 22:52
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Further, is there a way to see why without considering the idea of "dominated convergence"? – Math12345 Apr 22 '17 at 22:54
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Since you can view an infinite sum of functions as the limit of the sequence of partial sums, if the partial sums are all continuous and the convergence of of the sequence is uniform then the limit function will be continuous which will justify the limit and sum swap. All the facts stated here are quite standard analysis arguements, if you have any questions on how to prove them I can elaborate. However you can find any of these proofs online. Hope this helps!
ADA
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Just Dominated convergence suffices, no need of uniform convergence:see here – vidyarthi Apr 19 '17 at 05:32
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Of course that will work, but I am not sure the OP knows anything about measure theory. – ADA Apr 19 '17 at 05:36
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ok, but I think your writing "you can find any of these proofs online" needs a modification, though. – vidyarthi Apr 19 '17 at 05:40
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I mean that it is not easy to obtain a direct reference to the OP's question, say, by a direct google search – vidyarthi Apr 19 '17 at 05:47