I asked on MSE this question which I am going to copy-paste here:


"In mathematics, the Riemann series theorem (also called the Riemann rearrangement theorem), named after 19th-century German mathematician Bernhard Riemann, says that if an infinite series of real numbers is conditionally convergent, then its terms can be arranged in a permutation so that the new series converges to an arbitrary real number, or diverges."

Question of mine is:

If $\sum a_k$ is some conditionally convergent series and $c \in (- \infty, + \infty)$ then is there at least a countably infinite number of permutations $\sigma_m(c);m=1,2,...$ such that $\sum a_{\sigma_m(c)}=c$ for every $c \in (- \infty, + \infty)$?"

There I received a very nice answer:

"Yes. Let the sum of the first $n$ terms of your series be $s_n$. Then the remainder of the series is conditionally convergent, so it can be rearranged to achieve a sum of $c-s_n$. Choose one, and then move on. For each $n$ this yields at least $n!$ permutations that give the desired sum, no more than $n$ of which can have been duplicated at a prior step."

I got an idea from the comment on that answer for a question to ask it here on MO, here it is:

For $\sum a_k$, which is some conditionally convergent series, the number of sums $\sum a_{\sigma}$, where $\sigma$ is some permutation, is uncountable. For every $c \in (-\infty, + \infty)$ there is the set $P_c$ of all permutations $\psi(c)$ for which $\sum a_{\psi(c)}=c$. Can every $P_c$ be uncountable?


Every $P_c$ has the size of the reals.

For instance, suppose $\sum_n a_n=c$ and start by writing $\mathbb N=A\cup B$ where $\sum_{n\in A}a_n$ converges absolutely (to $a$, say). This is possible because $a_n\to 0$: Let $m_0<m_1<\dots $ be strictly increasing and such that for any $k$, $|a_n|<2^{-k}$ for all $n\ge m_k$, and then set $A=\{m_k:k\in\mathbb N\}$. It follows that $B$ is infinite and $\sum_{n\in B}a_n$ converges conditionally (to $c-a$).

Now, if $\pi$ is any permutation of $\mathbb N$, consider the permutation $\tau_\pi$ given by $\tau_\pi(n)=n$ if $n\in B$ and $\tau_\pi(m_k)=m_{\pi(k)}$ for all $k$. It is easy to check that $\tau_\pi\in P_c$.

  • $\begingroup$ We can choose positive $a$ and all $a_n$ positive without considerable loss of generality? Is this right? $\endgroup$ – user147968 Nov 1 '19 at 0:46
  • $\begingroup$ Yes, we can always pick them positive if we want. $\endgroup$ – Andrés E. Caicedo Nov 1 '19 at 1:01
  • $\begingroup$ But if you choose some $\sum_{n\in A}a_n$ then you have chosen also the indices $n \in A$ and then you are choosing that $|a_n|<2^{-k}$. How is this bounding possible if indices are already chosen before? $\endgroup$ – user147968 Nov 1 '19 at 1:09
  • $\begingroup$ I meant: choose the indices in $A$ and call $a$ the corresponding sum. That's the construction described in the answer. But, as you pointed out, we could instead have fixed $a$ first and then picked $A$ so the corresponding sum converges absolutely to $a$. Lots of flexibility in the argument $\endgroup$ – Andrés E. Caicedo Nov 1 '19 at 1:17
  • $\begingroup$ I understand much of your argument in your second comment, but even if you choose first the indices in $A$ how it is possible later to bound them like you mentioned: $|a_n|<2^{-k}$ ? Because, if first what you decide is to choose them, how can they later be bounded like that? $\endgroup$ – user147968 Nov 1 '19 at 1:21

Let $G$ be the set of all permutations $\sigma$ such that for all positive integers $n$, we have $\{\sigma(2n-1),\sigma(2n)\} = \{2n-1,2n\}$. There are as many elements of $G$ as there are subsets of the even numbers, so $\# G =\mathfrak{c} = 2^{\aleph_0}$ is the cardinality of all permutations on the positive integers. It is easy to see that if $\sum_{n=1}^{\infty} a_n = c$ then for all $\sigma \in G$ we have $\sum_{n=1}^{\infty} a_{\sigma(n)} = c$.

[Actually $G$ is a subgroup, isomorphic to $\prod_{n=1}^{\infty} \mathbb{Z}/2\mathbb{Z}$.]

Let $\sum_{n=1}^{\infty} a_n$ be a conditionally convergent real series. By Riemann's Theorem you have a permutation $\sigma_c$ such that $\sum_{n=1}^{\infty} a_n = c$. Therefore any permutation of the form $\sigma = \sigma_g \sigma_c$ for $\sigma \in G$ will also have $\sum_{n=1}^{\infty} a_{\sigma(n)} = c$. The number of these is $\# G$.

Remark: The precise condition on a permutation $\sigma$ of the positive integers that for all real convergent series $\sum_n a_n = c$ we have $\sum_n a_{\sigma(n)} = c$ was determined by Levi in 1946.

  • $\begingroup$ Pete, do you know of any reasonable notion of size that allows one to distinguish some of the $P_c$ from others (meaning, that some are "big")? This of course may depend on the specific series. (Measure and category do not seem to suffice.) $\endgroup$ – Andrés E. Caicedo Nov 1 '19 at 20:23
  • $\begingroup$ @AndrésE.Caicedo It seems possible that some asymptotic density could be defined to distinguish some of the $P_c$ from others. $\endgroup$ – user147968 Nov 2 '19 at 18:44

Serendipitously, I just happened to comment about this in Mathematics StackExchange 3 days ago. For various results along the lines of what you're asking, see this 18 September 2000 sci.math post.

The earliest publication I'm aware of in which this is at least implicitly shown is

Henry Lyon Trachtenberg [Trachtenburg] ($\approx$1883-??), On the proof of Riemann's theorem on semi-convergent series, Mathematical Gazette 2 #42 (December 1903), 361-362.

Trachtenberg begins by pointing out that proofs of the Riemann rearrangement theorem given by Riemann himself and given in Whittaker's 1902 book provide exactly one rearrangment that converges to a specified real number. In this paper Trachtenberg shows that, given a conditionally convergent series and a real number $S,$ there are infinitely many distinct rearrangements of the series that converge to $S.$ In fact, Trachtenberg's proof actually shows there are continuum many rearrangements that converge to $S$ (because, after looking at his proof, note that there are continuum many infinite sequences of positive integers), but he does not point this out. Trachtenberg also does not raise the question of whether uncountably many such rearrangements are possible. Trachtenberg's proof reappears (with more details included) on pp. 459-460 of Hobson's 1907 book, but neither Trachtenberg's name nor his publication is mentioned there.

I know of two earlier publications in which "infinitely many for each $S$" is also shown (see below), but the methods used seem to only show couuntably infinitely many.

Ulisse Dini (1845-1918), Sui prodotti infiniti [On infinite products], Annali di Matematica Pura ed Applicata (2) 2 #1 (August 1868), 28-38. [See pp. 28-31.]

Alfredo Capelli (1855-1910) and Giovanni Garbieri (1847-1931), Corso di Analisi Algebrica [Course of Algebraic Analysis], Volume Primo. Teorie Introduttorie [Volume One. Introductory Theory], F. Sacchetto (Padova, Italy), 1886, viii + 511 pages. [See p. 126, last sentence of #3]


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