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If a series converges absolutely, then it is known that the value of the series is independent of rearrangements. More precisely, if $\sum |a_n| < \infty$ and $\sigma:\Bbb N\to \Bbb N$ is a bijection then $\sum a_{\sigma(n)}$ converges and its value is independent of $\sigma$.

Now what about the converse? That is, if $\{a_n\}_{n=1}^\infty$ is an arbitrary real or complex sequence such that for every bijection $\sigma$ of $\Bbb N$, we have $\sum a_{\sigma(n)}$ converges to the same value, is it true that $\sum a_n$ coverges absolutely? I am interested in this in the context of signed measures on a measurable space à la Stein's and Shakarchi's definition of signed measures in their third volume on real analysis.

Alex Ortiz
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    If we're talking about sequences in some $\mathbb{R}^n$ or $\mathbb{C}^n$, then yes. Not in infinite-dimensional normed spaces. – Daniel Fischer Nov 16 '17 at 19:51
  • @DanielFischer Cool result! Could you elaborate? :) – Ant Nov 16 '17 at 19:53
  • @Ant For the finite-dimensional case, see Riemann. For the infinite-dimensional case, specific examples are easy (e.g. $a_n = \frac{1}{n} e_n$ in $\ell^p$ for $p > 1$). I think one can find an unconditionally convergent but not absolutely convergent series in every infinite-dimensional Banach space, but I don't know a proof off the top of my head. – Daniel Fischer Nov 16 '17 at 20:01
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    Ah, a theorem by A. Dvoretzky and C. A. Rogers. – Daniel Fischer Nov 16 '17 at 20:07
  • @DanielFischer The wiki article states the direction of the theorem that I say is known. I am wondering about the converse, which doesn't appear to be in that wiki article. – Alex Ortiz Nov 16 '17 at 20:14
  • Wikipedia defines conditional convergence as non-absolute convergence. So with that terminology, it states the other direction. – Daniel Fischer Nov 16 '17 at 20:23
  • @DanielFischer I posted an answer to my own question, but I am having some doubts about it. At first, I thought the contrapositive was as I wrote it, but now I suspect that it is actually "if $\sum a_n$ converges conditionally, then for each $\alpha\in\Bbb R$ there exists a bijection $\sigma$ such that $\sum a_{\sigma(n)} \ne \alpha$." Would you mind checking my answer to this question and possibly resolving this? – Alex Ortiz Nov 22 '17 at 21:57
  • Rather, "if $\sum a_n$ converges conditionally, then at least one [actually, both] of the following things happens: a) there is a permutation $\sigma$ such that $\sum a_{\sigma(n)}$ doesn't converge, and b) there are two permutations $\sigma,\tau$ such that $\sum a_{\sigma(n)} \neq \sum a_{\tau(n)}$ (and both of these series are convergent)." – Daniel Fischer Nov 22 '17 at 22:18

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Based on Daniel Fischer's comments, we can prove the converse is true by using the contrapositive. Breaking the statement down in the case of real series, we are asserting that if there exists $\alpha\in\Bbb R$ such that for every permutation $\sigma$ we have $\sum a_{\sigma(n)}$ converges and $\sum a_{\sigma(n)} = \alpha$, then $\sum a_n$ converges absolutely. The kind of convergence in the hypothesis is known as "unconditional convergence."

Thus, the contrapositive of the statement we're looking for is "if $\sum a_n$ is not absolutely convergent, then it is not unconditionally convergent."

If the series $\sum a_n$ is not absolutely convergent, then either $\sum a_n$ does not converge, or it converges, but not absolutely. In the first case, $\sum a_n$ is readily seen to be not unconditionally convergent, since unconditional convergence implies convergence in the usual sense. In the second case, $\sum a_n$ is convergent, but not absolutely, so the Riemann rearrangement theorem supplies two distinct permutations $\sigma,\tau$ such that $\sum a_{\sigma(n)},\sum a_{\tau(n)}$ both exist in the extended sense, possibly equal to $\pm\infty$, and they are unequal, so the series is not unconditionally convergent.

Alex Ortiz
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  • +1 I was slightly confused by this question because, while I knew Riemann's rearrangement theorem, it was always presented to me as "the reason why" we can rearrange an absolutely convergent series. Which of course it's not, since riemann only gives the converse direction (i.e. we cannot rearrange a conditional convergent series) but makes no statement about absolutely converging series. – Ant Nov 17 '17 at 07:49
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    Strictly speaking, the contrapositive of "unconditional convergence implies absolute convergence" is "If $\sum a_n$ is not absolutely convergent, then it is not unconditionally convergent". This is more general, since it contains the possibility that there is no rearrangement such that $\sum a_{\sigma(n)}$ converges (e.g. if $a_n$ doesn't converge to $0$, or $a_n = \frac{1}{n+1}$). But the only interesting case is when $\sum a_n$ is convergent, but not absolutely. For if $\sum a_n$ is not convergent, then it is trivially not unconditionally convergent. – Daniel Fischer Nov 22 '17 at 22:14
  • And if it is convergent but not absolutely, Riemann proved that it's not unconditionally convergent. – Daniel Fischer Nov 22 '17 at 22:14