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I know that for a (complex) series : If the series converge absolutely, then every rearrangement also converges and converge to the same limit.

But how to prove the converse : If every rearrangement of a (complex) series converge to the same limit, then the series itself also converge absolutely. That is to prove :

Let $\sum z_i$ be a complex series. If every rearrangement of $\sum z_i$ converge to the same limit, then $\sum z_i$ converges absolutely.

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  • $\begingroup$ No, the contrapositive is that if $\sum|z_i|$ does not converge, then either there is a divergent rearrangement, or there are two rearrangements converging to different limits. $\endgroup$ – Brian M. Scott Feb 17 '15 at 0:24
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The contrapositive of your statement follows from the Riemann series theorem.


To be more precise, suppose for a contradiction that $\sum z_i$ is a sum that is convergent under every rearrangement to the same limit $\ell$, but it is not absolutely convergent. Then $\sum|\Re z_i|+\sum|\Im z_i|\ge \sum|z_i|=\infty$, so either $\sum\Re z_i$ or $\sum\Im z_i$ is a conditionally convergent real series, so the Riemann series theorem gives a rearrangement $\sigma$ such that one of these two converges to something different from their original sum, and then $\sum z_{\sigma(i)}=\sum\Re z_{\sigma(i)}+\sum\Im z_{\sigma(i)}\ne\ell$ in contradiction to the hypothesis.

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    $\begingroup$ That theorem is for real series, so it doesn’t directly apply here. $\endgroup$ – Brian M. Scott Feb 17 '15 at 0:22
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    $\begingroup$ @BrianM.Scott I added more details. (You can also use the "Steinitz's theorem" extension mentioned on wiki to get the complex version directly.) $\endgroup$ – Mario Carneiro Feb 17 '15 at 0:40

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