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Suppose $\sum a_n$ is a series of complex numbers, if $\sum a_n$ and its every rearrangement all converge to the same sum, does $\sum a_n$ converge absolutely?

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  • $\begingroup$ After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?. $\endgroup$ – Clement C. Sep 11 '17 at 15:11
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Yes, since $\mathbb{C}$ has finite dimension. You may find this question helpful; quoting from the answer:

Theorem. (Dvoretsky-Rogers, 1950) Every unconditionally convergent series in a Banach space $X$ is absolutely convergent if and only if $X$ is finite dimensional.


To be clear: the above theorem is good to know, but overkill in your case: you want only the "easy" direction of this theorem. Here it goes:

  1. Since $\sum_n a_n$ converges unconditionally, then so do $\sum_n \operatorname{Re}(a_n)$ and $\sum_n \operatorname{Im}(a_n)$.

  2. These two are real-valued series, so by the result on $\mathbb{R}$ (which is given by (the contrapositive of) Riemann's rearrangement theorem) $\sum_n \operatorname{Re}(a_n)$ and $\sum_n \operatorname{Im}(a_n)$ are both absolutely convergent.

  3. Now, by the triangle inequality this immediately implies that $\sum_n a_n$ is absoltuely convergent as well: indeed, $$\sum_n \lvert a_n\rvert \leq \sum_n \lvert \operatorname{Re}(a_n)\rvert+\sum_n \lvert \operatorname{Im}(a_n)\rvert < \infty$$

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  • $\begingroup$ Is there an elementary proof for $\mathbb C$ ? $\endgroup$ – Gabriel Romon Sep 6 '17 at 12:18
  • $\begingroup$ Yes, I reckon. (1) If $\sum_n a_n$ converges unconditionally, then so do $\sum_n \operatorname{Re}(a_n)$ and $\sum_n \operatorname{Im}(a_n)$; (2) by the result on real-valued series, then $\sum_n \operatorname{Re}(a_n)$ and $\sum_n \operatorname{Im}(a_n)$ converge absolutely; (3) by the triangle inequality, this implies $\sum_na_n$ converges absolutely as well. @LeGrandDODOM $\endgroup$ – Clement C. Sep 6 '17 at 12:22
  • $\begingroup$ Thank you, and what's the gist of the proof for $\mathbb R$ ? $\endgroup$ – Gabriel Romon Sep 6 '17 at 12:26
  • $\begingroup$ To be clear: the "hard" direction in the theorem above is to prove the converse (which the OP didn't ask about, but that I figured you be nice for them to know): if every unconditionally convergent series is convergent, then the space has finite dimension. The other direction is easy. $\endgroup$ – Clement C. Sep 6 '17 at 12:26
  • $\begingroup$ @LeGrandDODOM Erm. That where the word "easy" in my comments above is misleading. AFAIK, the base case for $\mathbb{R}$ is (the contrapositive of) Riemann's rearrangement theorem. So... "easy" assuming that one. $\endgroup$ – Clement C. Sep 6 '17 at 12:28

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