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Jan 29, 2022 at 17:19 comment added Yaakov Baruch @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.
Jan 25, 2022 at 2:51 history edited Michael Hardy CC BY-SA 4.0
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Apr 13, 2017 at 12:19 history edited CommunityBot
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Aug 31, 2015 at 15:20 comment added Jérôme JEAN-CHARLES 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 $a_{f(n)}$.
Aug 14, 2015 at 12:54 answer added David E Speyer timeline score: 24
Aug 6, 2015 at 2:08 comment added Timothy Chow 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.
Aug 5, 2015 at 5:30 comment added Michael Hardy Levi's duality: researchgate.net/publication/… ${}\qquad{}$
Aug 5, 2015 at 5:23 comment added Michael Hardy 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,
Aug 5, 2015 at 5:20 comment added Michael Hardy $\ldots\,{}$ and another: projecteuclid.org/euclid.pjm/1102688295 ${}\qquad{}$
Aug 5, 2015 at 5:18 comment added Michael Hardy I'd better record the following here before I lose track of it, even if I don't look at it before tomorrow: link.springer.com/article/10.2478%2Fs11533-012-0156-x#page-1 ${}\qquad{}$
Aug 5, 2015 at 3:12 history edited Ricardo Andrade
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Aug 5, 2015 at 1:19 comment added Brendan McKay mathoverflow.net/questions/213064/… may (or may not) be interesting.
Aug 5, 2015 at 0:56 comment added Brendan McKay 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.
Aug 5, 2015 at 0:47 comment added Michael Hardy I also find this: "Rearrangements that Preserve Convergence", Journal of the London Mathematical Society, volume s2-15, issue 1, pages 134-142. jlms.oxfordjournals.org/content/s2-15/1/134.full.pdf+html ${}\qquad{}$
Aug 5, 2015 at 0:38 comment added Michael Hardy 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${}\,\ldots$". ${}\qquad{}$
Aug 4, 2015 at 23:24 comment added Gerry Myerson Also, Garibay et al., The geometry of sum-preserving permutations, Pac J Math 135 (1988) 313-322, MR0968615 (90f:40001).
Aug 4, 2015 at 23:16 comment added Gerry Myerson See Paul Schaefer, Sum-preserving rearrangements of infinite series, Amer Math Monthly 88 (1981) 33-40.
Aug 4, 2015 at 20:48 history asked Michael Hardy CC BY-SA 3.0