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Define $S_n = \sum_{j=1}^n a_j$ a sum of real numers.

Suppose there is a subsequence such that $S_{n_j} \to S$ for some $S$, i.e., the infinite sum converges along this subsequence. Does this imply that $S_n$ converges too?

It seems like it should, since it's simply an index which is increasing but there may be a counterexample I can't think of..

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    $\begingroup$ If all the terms $a_j$ are positive, then it is true. Since $S_n$ is then an increases sequence which converges if and only if it is bounded, if and only if a subsequence is bounded (since it's increasing). $\endgroup$ – breeden Jun 24 '15 at 17:47
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No, if e.g. $a_j = (-1)^j$ then $S_{2k} = 0$ for all $k$ but the series itself does not converge.

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  • $\begingroup$ If it is Cauchy I think it does. $\endgroup$ – user123124 Jun 24 '15 at 17:51
  • $\begingroup$ @Johan The sequence $S_n$ is not cauchy. It's $-1$ for odd indexes, and $0$ for even indexes. Cauchy sequence must have elements that get close to each other. $\endgroup$ – muaddib Jun 24 '15 at 17:53
  • $\begingroup$ @maddib I know it was just a remark to the case when it holds. $\endgroup$ – user123124 Jun 24 '15 at 17:54
  • $\begingroup$ @Johan Understood. $\endgroup$ – muaddib Jun 24 '15 at 17:55
  • $\begingroup$ @Johan do you mean when $a_n$ are cauchy? Because even that doesn't work (if you are only asking about arbitrary subsequencs), you can just take $a_j$ to be $n^2$ copies of $1/n^2$ and then $n^2$ copies of $-1/n^2$, etc... [edit: Actually, now I think about it, I think you meant when $S_n$ are cauchy and you have a convergent subsequence, which is true] $\endgroup$ – breeden Jun 24 '15 at 17:56

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