# Does an unconditionally convergent series of complex numbers converge absolutely?

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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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, $$\sum_n \lvert a_n\rvert \leq \sum_n \lvert \operatorname{Re}(a_n)\rvert+\sum_n \lvert \operatorname{Im}(a_n)\rvert < \infty$$

• Is there an elementary proof for $\mathbb C$ ? – Gabriel Romon Sep 6 '17 at 12:18
• 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 – Clement C. Sep 6 '17 at 12:22
• Thank you, and what's the gist of the proof for $\mathbb R$ ? – Gabriel Romon Sep 6 '17 at 12:26
• 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. – Clement C. Sep 6 '17 at 12:26
• @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. – Clement C. Sep 6 '17 at 12:28