There's a well-known result that if $\sum a_n$ is conditionally convergent, then for any real $c$ there exists a permutation $\pi:\mathbb{N} \to \mathbb{N}$ such that $\sum a_{\pi(n)} = c$.

A consequence of this is that you cannot get away with rearranging infinite sums, in general. However, if the rearrangement is sufficiently tame, then we can get away this. For instance, if we rearrange only finitely many terms, we can get away with this without the value of the sum changing. Indeed, there is a stronger result that if we have a permutation $\pi$ and a uniform constant $M$ such that $|\pi(n) - n| < M$ for all positive integers $n$, then $\sum a_n = \sum a_{\pi(n)}$. This is not too hard to prove. My question is, essentially, can we get a "maximally strong" result of this sort?

Specifically I ask the following question:

Characterize permutations $\pi:\mathbb{N} \to \mathbb{N}$ with the following property:

For any convergent infinite sum $\sum a_n$, we have that $\sum a_n = \sum a_{\pi(n)}$.

Notice the order of quantifiers here.

It's conceivable that the result I described in the second paragraph is the best we can do, in the sense that if $\sup_n |\pi(n) - n| = \infty$, there exists a conditionally convergent $a_n$ with $\sum a_n \neq \sum a_{\pi(n)}$. Perhaps we can do better, though. One guess is slightly changing the hypotheses as follows: there exists a uniform $M$ such that $|\pi(n) - n| < M$ for all $n \in \mathbb{N} \setminus S$, where $S$ is a subset of $\mathbb{N}$ with $0$ asymptotic density.

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    $\begingroup$ How about splitting up $\Bbb N$ into a sequence of intervals increasing in length and then applying a cyclic permutation to each of those intervals? $\endgroup$ – Lord Shark the Unknown Dec 13 '18 at 7:47
  • $\begingroup$ @LordSharktheUnknown Indeed, that would refute the claim in the first sentence of the last paragraph above. But are you sure this is always going to work? $\endgroup$ – MathematicsStudent1122 Dec 13 '18 at 7:51
  • $\begingroup$ See my comments and answer to this question. I don't know to what extent it answers your question(s), but it certainly should give you a starting point, although maybe you've seen this since I posted it fairly recently. (I'm about to participate in a Skype meeting and can't look at your question in any more detail now.) $\endgroup$ – Dave L. Renfro Dec 13 '18 at 14:52
  • $\begingroup$ The answer seems to be "There exists $M>0$ such that for every $n$, the set $\{\pi(1),\dots,\pi(n)\}$ is a union of at most $M$ intervals of consecutive integers". $\endgroup$ – fedja Dec 13 '18 at 15:08
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    $\begingroup$ @MathematicsStudent1122 : In the Lord Shark idea of cyclic permutations over intervals, at every step, the re-ordered sum differs in at most one term from the original sum order. So the sums differ by something that looks like $a_i-a_j$, where $i,j\rightarrow\infty$. $\endgroup$ – Michael Dec 13 '18 at 15:41

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