Rearrangement, conditional convergence, and "placid" permutations This question came out of a conversation with my students about Riemann's rearrangement theorem, and the general problem of which permutations are "safe" w/r/t summing infinite series. It was previously asked and bountied at MSE without success.
Let $S_\infty$ be the group of permutations of $\mathbb{N}$. For a sequence $\mathscr{A}=(a_i)_{i\in\mathbb{N}}$, say that a permutation $p\in S_\infty$ is $\mathscr{A}$-placid iff for every $q\in S_\infty$ and every pair of integers $z_0, z_1$ we have $$\sum_{i\in\mathbb{N}}a_{q(i)}\simeq \sum_{i\in\mathbb{N}}a_{p^{z_0}\circ q\circ p^{z_1}(i)}$$ where "$s\simeq t$" means "either $s$ and $t$ are each undefined, or they are defined and equal." Basically, $p$ is $\mathscr{A}$-placid if $p$ is never interesting from the point of view of rearranging the terms in $\mathscr{A}$. For example, the permutation swapping $2i$ and $2i+1$ for each $i$ is $\mathscr{A}$-placid for every $\mathscr{A}$.
I'm curious whether placidity actually depends on the sequence in question (restricting attention to sequences whose corresponding series converge conditionally, to avoid triviality). My main question is the following:

Question 1: Are the following equivalent?

*

*$p$ is placid with respect to the alternating harmonic sequence $((-1)^{i+1}{1\over i})_{i\in\mathbb{N}}$.


*$p$ is $\mathscr{A}$-placid for every conditionally convergent sequence $\mathscr{A}$.

I suspect the answer is negative, but I don't immediately see how to prove it. Incidentally, as far as I can tell the only "obviously placid" permutations are those in which there is a finite bound on the distance an element of $\mathbb{N}$ is moved. Merely having finite orbits isn't enough: for example, for any conditionally convergent sequence $\mathscr{S}$ there is a permutation of $\mathbb{N}$ of order $2$ which applied to $\mathscr{S}$ results in a series with limit $+\infty$. This was pointed out to me by a colleague after I brashly claimed otherwise!
EDIT: Since this question looks like it might be more complicated than I originally thought, I'll add a second, hopefully easier aspect:

Question 2: Are there conditionally convergent series $\mathscr{A},\mathscr{B}$ such that "$\mathscr{A}$-placid" is not the same as "$\mathscr{B}$-placid" (that is, is the notion of placidity nontrivial in the first place)?

Of course a negative answer to question 1 would give a positive answer to question 2.
 A: As a preamble, let me quickly address this comment from the original question:

Incidentally, as far as I can tell the only "obviously placid" permutations are those in which there is a finite bound on the distance an element of N is moved.

A simple example of a placid permutation with unbounded distances moved is the one that reverses the intervals $[1, 2]; [3, 6]; [7, 12]; [13, 20]; \dotsc$, ie the involution
$$ 1\leftrightarrow 2; 3\leftrightarrow 6, 4 \leftrightarrow 5; 7\leftrightarrow 12, 8\leftrightarrow 11, 9\leftrightarrow 10; 13 \leftrightarrow 20, 14 \leftrightarrow 19, \dotsc. $$

I believe that the answer to the first question is positive and the second question is negative.  That is, the notion of placidity doesn't depend on the particular conditionally convergent series. I am not 100% sure of the details, but I hope to provide a sketch here:
A classical theorem of Agnew (and Levi and Pleasants; see also Guha and Schaefer) characterizes which permutations preserve convergence and sums of conditionally convergent series.
Theorem (Agnew).  A permutation of the natural numbers $\pi$ preserves the sum of all conditionally convergent series if and only if there is an integer $M$ such that the image of every initial prefix of $\mathbb N$ is a union of $M$ or fewer intervals of $\mathbb N$.
This characterization allows for some divergent series to be rearranged into a convergent series, whereas your definition has an "iff" constraint, so for your purposes this property must be satisfied by both $\pi$ and $\pi^{-1}$.
The proof of the theorem is relatively simple.  In one direction, one finds that every partial sum of the rearranged series can be written as an alternating sum of a bounded number of partial sums of the original series, and so by a Cauchy-type criterion, the rearranged series converges to the same sum.  In the other direction, one constructs an explicit (conditionally) convergent series whose sum is "ruined" by the permutation.  Let's look at the specific proof of Theorem 3 given starting on page 13 (14 of 38 in the PDF) in the masters thesis "On rearrangements of conditionally convergent series" by ‪Balázs Gerencsér.
For any $\pi$ not satisfying the condition above, Gerencsér constructs a series whose convergence is ruined.  That series consists of a rearrangement of infinitely many zeroes and $i$ copies each of $1/\sqrt{i}$ and $-1/\sqrt{i}$ for each positive integer $i$.  In particular, Gerencsér demonstrates the placidity for a single specific series fully captures the notion of preserving the sum of all conditionally convergent series.
I think that essentially the same proof may be obtained by rearranging any conditionally convergent series, but the details are a bit messy.  I will refer directly to notation from that proof.  The proof proceeds in steps indexed by integers $i\ge 2$ and constructs a series $a_j$ with two main properties:

*

*The partial sums of $a_j$ are small in absolute value: $\lvert\sum_{k=1}^{n}a_k \rvert\le 1/\sqrt{i-1}$ if $n\ge N_i$.


*The partial sums of the rearranged $a_j$ "constructively interfere": $\lvert\sum_{k=1}^{N_i} a_{\pi(k)}\rvert \ge \sqrt{i-1}$.
Suppose we attempt the same strategy by rearranging an arbitrary conditionally convergent series $a_j$.  We run into a few difficulties relative to the series rearranged by Gerencsér.  One issue that that the terms might be "larger" than $1/\sqrt{i}$ and so condition 1 is hard to achieve.  On the other hand, the terms might be "smaller" than $1/\sqrt{i}$ and so it could be hard to achieve condition 2 as strongly.  But it should all work out, perhaps with a proliferation of indices and notation.
Fix a permutation $\pi$ that does not satisfy Agnew's condition.  (Switch $\pi$ and $\pi^{-1}$ if necessary.)  Fix a conditionally convergent series $a_k$.  We will construct a permutation $\rho$ so that $\sum a_{\rho(k)}=0$ but $\sum a_{\pi(\rho(k))} = +\infty$.  We proceed in steps as before, indexed by positive integers $\ell$.  At each point, we will have a "partial" permutation in the sense of partial function.  Each step consists of two sub-steps:

*

*First, consider all $k$ so that $\lvert a_k\rvert \ge 1/\ell$.  There are finitely many such $k$, so they are all below some bound $K_\ell$.  Possibly by increasing $K_\ell$, define $\rho$ on all $k\le K_\ell$ so that $\lvert \sum_{k=1}^{n} a_{\rho(k)}\rvert \le 1/\ell$ for all $K_{\ell-1}\le n\le K_\ell$.  Moreover, we want $\lvert \sum_{k=1}^{K_\ell} a_{\rho(k)}\rvert \le 1/(\ell+1)$ so that we can keep going in the next step.  This is done by the usual way in Riemann's original theorem on conditionally convergent series.  Essentially, occasionally you "use up" a positive term $a_k$ and then balance it by some number of negative terms; OR, you "use up" a negative term $a_k$ and then balance it by some number of positive terms.

The purpose of this sub-step is the same as the first property above: ensure that initial prefixes of $\sum a_{\rho(k)}$ are small.  Moreover, they decrease to $0$ and so $\sum a_{\rho(k)}=0$.  Naturally, we must have another sub-step that has the opposite effect:


*We must "ruin" $\sum a_{\pi(\rho(k))}$.  To do so, we consider the $N_i$ from Gerencsér's proof.  The strategy is the same, but I find it very hard to write down all of the notation for it.

In Gerencsér's proof, one "just" chooses $a_{j_k-1}$ to be big and $a_{j_k}=a_{j_k-1}$ to cancel it directly which allows the preservation of all of the desired structure.  If we are rearranging an arbitrary series, then the issue is that we may not have enough "big" terms available in the right quantities to make this happen directly.  But something like the following should work:
Consider the terms of $a_k$ that have not yet been rearranged.  Keep picking positive terms until their sum exceeds $1/\ell$, then keep picking negative terms until their sum dips below zero.  Do this $\ell^2$ times.  Suppose $L$ terms were rearranged in total.  The idea is that partial sums of this rearrangement stay "small" (less than $1/\ell$), but if you consider only the positive terms, the sum exceeds $\ell$.  Then the idea is to use the positive terms as an analogue of $a_{j_k-1}$ and the negative terms as an analogue of $a_{j_k}$.
We can't achieve this directly by just assigning individual entries of $\rho$, but we can assign entire blocks.  In order to do so using the notation of Gerencsér's proof, I think it should suffice to consider $i=L$ and work with those indices in the proof.
I apologize by cutting this sketch off at this point, but I'm really struggling with the notation and I must go do something else.  Please let me know if this sketch makes sense or if you want some more/cleaner details.  I should be able to answer questions later.
References:

*

*Ralph Agnew. "Permutations preserving convergence of series".  Proceedings of the American Mathematical Society (1955).

*Balázs Gerencsér. "On rearrangements of conditionally convergent series".  M. Sc. Thesis from Eötvös Loránd University, Institute of Mathematics (2007).

*U. C. Guha, "On Levi's theorem on rearrangement of convergent series". Indian Journal of Mathematics (1967).

*F. W. Levi, "Rearrangement of convergent series". Duke Math Journal (1946).

*P. A. B. Pleasants, "Rearrangements that preserve convergence". Journal of the London Mathematical Society (1977)

*Paul Schaefer, "Sum-preserving rearrangements of infinite series". American Mathematical Monthly (1981).

A: Some tentative thoughts. (EDIT2) Following the mention of Agnew's theorem by @aorq and the good example he gives, I see the issue is not only a matter of how far the permutation displaces elements, but also a matter of how blocks of joint elements are dismantled. So I delete what I wrote under "I am inclined (without proof) to think that", which was plain wrong. (end EDIT2)
A remark to begin with: for a sequence $\mathcal{A}$, a permutation $p$ is $\mathcal{A}$-placid iff $p$ is $\mathcal{B}$-placid for any $\mathcal{B}$ that is a permutation of $\mathcal{A}$.
So we can deal with sequences $\mathcal{A}$ where absolute value of $a_n$ is decreasing, and then generalize placidity results to sequences which are permutations of those sequences: this includes any sequence that has no zero term.
Some definitions:

*

*For a permutation $p$, a sequence $\mathcal{A}$, and $n \in \mathbb{R}$, let $s(N)$ be the number of $n \le N$ such that $p(n) \gt N$.


*Let $q(N)$ be $\min \{n,  n \le N \land p(n) > N\}$. All $n < q(N)$ stay below $N$ when permuted, i.e. $n < q(N) \implies p(n) < N$.


*Let's call $a'_n = a_{p^{-1}(n)}$, which is equivalent to $a'_{p(n)} = a_n$.
We can spot two cases, one where the two series are equivalent, one where they are not:

*

*If $s(N) |a_{q(N)}| \to 0$ when $N \to \infty$,
$|\sum_{n=0}^N a_n - \sum_{n=0}^N a'_n| \le 2 s(N) |a_{q(N)}|$.
This because $a_n$ is supposed ordered by decreasing absolute value.
As $s(N) |a_{q(N)}| \to 0$, $\sum a_n \simeq \sum a'_n$.


*If $s(N) |a_N| \nrightarrow 0$ when $N \to \infty$, $\exists \varepsilon > 0, \forall N, s(N) |a_N| \ge \varepsilon $, we first apply a permutation to $\mathcal{A}$ such that we have a sequence $(M_i)_{i \in \mathbb N}$ where all the $a_n$ that permutation $p$ throws away from $[0, M_i]$ to $]M_i, \infty[$ are positive terms.
Then $|\sum_{n=0}^N a_n - \sum_{n=0}^N a'_n| \ge s(N) |a_N| \ge \varepsilon$.
So the two series are either non-convergent, or convergent to values separated by at least $\varepsilon$.
Missing step 1: to prove that those conditions (either $s(N) |a_{q(N)}| \to 0$ when $N \to \infty$, or $s(N) |a_N| \nrightarrow 0$ when $N \to \infty$) are maintained when iterating $p$, i.e. replacing $p$ by $p^k$.
(EDIT) Missing step 2: to prove that those conditions are maintained when applying any permutation to sequence $\mathcal{A}$.
This may in fact be wrong, for the following reason. We have seen above that permutations have more effect on slowly decreasing (in absolute value) sequences than on quickly decreasing ones. However, it is possible to apply a permutation to a quickly decreasing sequence $\mathcal{A}$, that groups together a sufficient number of same_sign terms, so that the sum of blocks of same-sign terms make a slowly decreasing sequence $\mathcal{B}$. For example, given $\sum (-1)^n/n$, it is possible to group together some positive and then some negative terms such that some partial sums approximate those of $\sum (-1)^n/\sqrt n$. Of course this is not a proof, because the number of terms in a block increases to $\infty$, so we cannot say that a permutation on sequence $\mathcal{B}$ will have the same effect than a permutation on sequence $\mathcal{A}$. (end EDIT)
This also leaves a gap between $s(N) |a_{q(N)}| \to 0$ when $N \to \infty$ and $s(N) |a_N| \nrightarrow 0$ when $N \to \infty$, where nothing can be said.
