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The standard rearrangement theorem for conditionally convergent series says that the terms in a conditionally convergent series can be rearranged so that the new sum is any desired number, or $\pm\infty$.

Let me set up some notation to investigate this a little further. My coefficients will be determined by a function $\alpha:\mathbb{N}\to\mathbb{R}$ (or $\mathbb{C}$). Then I'll simply write $\sum\alpha$ for this series $\sum_{n=1}^\infty \alpha(n)$.

Now we let $\operatorname{Sym}(\mathbb{N})$ act on the set $A = \{ \alpha: \mathbb{N}\to \mathbb{R} \}$ by the rule $\alpha^\sigma (x) = \alpha(\sigma(x))$. Then we can ask simple questions

  1. is $\sum \alpha^\sigma = \sum \alpha$?
  2. if $\sum\alpha$ converges, does $\sum\alpha^\sigma$ also converge?

But these are not really the questions I want to ask.

Let $B\subseteq A$ be the subset of all sequences $\alpha$ so that the series $\sum\alpha$ is conditionally convergent.

Question 1 What is the stabilizer $G = \{ \sigma \mid \alpha^\sigma \in B\ \forall \alpha\in B\}$?

Question 2 What is the group of sum-preserving permutations $H = \{ \sigma \mid \sum \alpha^\sigma = \sum\alpha \ \forall \alpha\in B\}$?

Certainly any permutation which is the identity off a finite subset of $\mathbb{N}$ is contained in both.

There are some subgroups of $\operatorname{Sym}(\mathbb{N})$

This is related to the question here.

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    $\begingroup$ If there is a constant $C$ such that $|\sigma(n)-n| \leq C$ for all $n$, then $\sigma$ is sum-preserving. And this "boundedness" property is preserved by compositions and inverses. So one might guess that perhaps these are all of the sum-preserving permutations. But it is not the case. There are sum-preserving permutations that have "unbounded steps" (but, I suppose, very far apart). There are several known characterizations of sum-preserving permutations, see for example projecteuclid.org/euclid.pjm/1102688295 and references therein. $\endgroup$
    – Zach Teitler
    Commented Mar 1, 2017 at 5:15
  • $\begingroup$ Maybe I'm blind, but why are G and H even groups? $\endgroup$
    – Johannes Hahn
    Commented Jun 20, 2017 at 21:18
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    $\begingroup$ There's a finer sufficient condition for being sum-preserving given in Bourbaki, TG (=Topologie Générale), IV, §7, exercise 12 (page IV.60 in the French edition), namely that $r(n) := |\sigma(n)-n|\cdot\sup_{m\geq n}|\alpha(m)|$ tends to $0$. I seem to remember there's a slight mistake and that it should be something different (maybe replace some $m$ or $n$ by $\sigma(m)$ or $\sigma(n)$), I guess solving the exercise will unravel the correct version. $\endgroup$
    – Gro-Tsen
    Commented Jun 20, 2017 at 22:20

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[Making comment into answer.] If there is a constant $C$ such that $|\sigma(n)−n| \leq C$ for all $n$, then $\sigma$ is sum-preserving. And this "boundedness" property is preserved by compositions and inverses. So one might guess that perhaps these are all of the sum-preserving permutations. But it is not the case. There are sum-preserving permutations that have "unbounded steps" (but, I suppose, very far apart). There are several known characterizations of sum-preserving permutations, see for example http://projecteuclid.org/euclid.pjm/1102688295 and references therein.

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