Rearrangements that never change the value of a sum

Question

I posted this question on math.stackexchange.com and so far the only answer posted (also mentioned in the comments under the question) shows that one of my rash initial guesses about the bottom-line answer was wrong.

Which bijections $f:\{1,2,3,\ldots\}\to\{1,2,3,\ldots\}$ have the property that for every sequence $\{a_n\}_{n=1}^\infty$, $$ \lim_{n\to\infty} \sum_{k=1}^n a_k = \lim_{n\to\infty} \sum_{k=1}^n a_{f(k)}, $$ where "$=$" is construed as meaning that if either limit exists then so does the other and in that case then they are equal?

Here's another rash initial guess, different from the one I posted on stackexchange: It's the bijections whose every orbit is finite.

(That there are uncountably many such bijections can be seen as follows: For each odd number $n$, let $f$ either fix $n$ and $n+1$ or interchange them. That bijection never changes the values of sums. It's a countably infinite sequence of binary choices, so it's uncountable.)

See Paul Schaefer, Sum-preserving rearrangements of infinite series, Amer Math Monthly 88 (1981) 33-40. — Gerry Myerson, Aug 04 '15 at 23:16
Also, Garibay et al., The geometry of sum-preserving permutations, Pac J Math 135 (1988) 313-322, MR0968615 (90f:40001). — Gerry Myerson, Aug 04 '15 at 23:24
What made me think of this question is that I have a sequence $A_1\subseteq A_2 \subseteq A_3\subseteq\cdots\subseteq\mathbb N$ and I want to say that $\lim\limits_{n=1}\sum\limits_{k\in A_n} a_k$ is not altered by reshuffling the members of $A_{n+1}\setminus A_n$ for any or all values of $n$ and I'd like to write something between the extremes of "Obviously${},\ldots$" and "Here is a proof${},\ldots$". ${}\qquad{}$ — Michael Hardy, Aug 05 '15 at 00:38
I also find this: "Rearrangements that Preserve Convergence", Journal of the London Mathematical Society, volume s2-15, issue 1, pages 134-142. http://jlms.oxfordjournals.org/content/s2-15/1/134.full.pdf+html ${}\qquad{}$ — Michael Hardy, Aug 05 '15 at 00:47
Your conjecture cannot be true since the set of permutations with only finite cycles is not closed under multiplication. Consider $g$ = (1 2)(3 4)(5 6)... and $h$ = (2 3)(4 5)(6 7)... . A better conjecture would be to allow all finite products of such permutations. Whether it is the same as the answer of Garibay et al that Gerry mentioned is another question. — Brendan McKay, Aug 05 '15 at 00:56
http://mathoverflow.net/questions/213064/a-collatz-like-question-about-permutations may (or may not) be interesting. — Brendan McKay, Aug 05 '15 at 01:19
I'd better record the following here before I lose track of it, even if I don't look at it before tomorrow: http://link.springer.com/article/10.2478%2Fs11533-012-0156-x#page-1 ${}\qquad{}$ — Michael Hardy, Aug 05 '15 at 05:18
$\ldots,{}$ and another: http://projecteuclid.org/euclid.pjm/1102688295 ${}\qquad{}$ — Michael Hardy, Aug 05 '15 at 05:20
Agnew R.P., Permutations preserving convergence of series, Proc. Amer. Math. Soc., 1955, 6(4), 563–564, Levi F.W., Rearrangement of convergent series, Duke Math. J., 1946, 13, 579–585, Pleasants P.A.B., Rearrangements that preserve convergence, J. London Math. Soc., 1977, 15(1), 134–142, Schaefer P., Sum-preserving rearrangements of infinite series, Amer. Math. Monthly, 1981, 88(1), 33–40, — Michael Hardy, Aug 05 '15 at 05:23
Levi's duality: http://www.researchgate.net/publication/31336855_On_Levi%27s_Duality_Between_Permutations_and_Convergent_Series ${}\qquad{}$ — Michael Hardy, Aug 05 '15 at 05:30
Gerry Myerson already mentioned Schaefer's paper. By the way, as a matter of etiquette, I'd recommend that next time you do all of this literature searching ahead of time, before posting your question to MO, rather than using MO as a scratchpad. — Timothy Chow, Aug 06 '15 at 02:08
A necessary and sufficient condition is $| {n - f(n) } | $ is bounded. Otherwise you can makeup a serie with $0$ for sum of $a_n$ and infinity for sum of $a_{f(n)}$. — Jérôme JEAN-CHARLES, Aug 31 '15 at 15:20
@JérômeJEAN-CHARLES. Boundedness easily implies equal sums. The other way is not true. Consider $f(n)=n+1$ if $n$ is not square, and $f(n^2)=(n-1)^2+1$. $|n-f(n)|$ is unbounded, but it's easy to see that the 2 partial sums up to $N$ always differ at most by 2 terms of the sequence, with indices going to $\infty$ with $N$, therefore their limits are equal. — Yaakov Baruch, Jan 29 '22 at 17:19

score 24 · Answer 1 · edited Jun 20 '17 at 07:45

For the purpose of recording an answer rather than just a pile of links: Michael Hardy requires that

if either limit exists then so does the other and in that case then they are equal?

Let's call this set $G$.

Levi answered a slightly different question, namely characterizing the permutations for which

if the left hand limit exists then so does the right and in that case then they are equal?

I'll call that $P$. Clearly, $G = P \cap P^{-1}$.

Theorem A permutation $f$ is in $P$ if and only if there is a constant $M$ such that, for any $N$, the set $f([1,N])$ can be written as $\bigcup_{i=1}^M [a_i, b_i]$, where $[a,b] = \{ a,a+1, \ldots, b \}$.

Levi provided a different characterization than this one; Agnew provided this characterization; I learned about both from Schaefer who points out that they are fairly directly equivalent.

Pleasants shows that $P$ is not closed under inversion. I haven't (in an one hour skim) found any papers which give a simpler characterization of $P \cap P^{-1}$ than the defining formula.

Remark I would find it nicer to restate Levi/Agnew's characterization as follows: For $S \subseteq \mathbb{N}$, define the blockiness of $S$ to be the least integer $\beta(S)$ such that $S$ can be written as $\bigcup_{i=1}^{\beta(S)} [a_i, b_i]$. (We could have $\beta(S) = \infty$.) Then $f$ is in $P$ if and only if there is a constant $M$ such that $\beta(f(S)) \leq M \beta(S)$. This makes it more obvious that $P$ is closed under composition.

Rearrangements that never change the value of a sum

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