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Suppose that $\sum^\infty_{n=1} u_n = s,$ where the series converges conditionally, and $s'>s$. How to prove the existence of such a permutation $\sigma,$ such that 1) $u_n\geq 0 \rightarrow \sigma(n)=n$ 2) $\sum^\infty_{n=1} u_{\sigma(n)} = s'?$

I know how to rearrange positive divergent series to get speed of divergence as low as needed, but I don't know, how to apply this to mentioned problem

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Have you seen en.wikipedia.org/wiki/Riemann_series_theorem already? –  J. M. Nov 28 '10 at 15:51
This site is intended for research-level mathematics. The FAQ lists a number of other sites (eg math.stackexchange.com) which are more appropriate for lower-level questions like this. FYI, your question is answered in Rudin's "Principles of mathematical analysis". –  Andy Putman Nov 28 '10 at 15:54
The statement I wrote about isn't a Riemann's theorem, which is well known. I consider only the permutations of NEGATIVE part of the series. In this case the proof of Riemann's theorem fails. –  Georgii Nov 28 '10 at 17:33
This extension of the Riemann Rearrangement Theorem is due to W. Sierpinski. It was established in a 1911 paper, Sur une propriété des séries qui ne sont pas absolument convergentes. From what I recall the proof is indeed more involved than the usual rearrangement argument. (I am not quite sure why the question was closed.) –  Pete L. Clark Nov 28 '10 at 23:28
Whoops, there were several votes to close already so I didn't read the question as closely as I should have. Sorry! –  Andy Putman Nov 29 '10 at 2:31

1 Answer 1


Since he provided the reference to the original proof, I'll let Pete L. Clark address Sierpiński's original result.

Instead, I'll add some brief remarks. In a series of three papers:

a. Wacław Sierpiński, "Contribution à la théorie des séries divergentes", Comp. Rend. Soc. Sci. Varsovie 3 (1910) 89–93 (in Polish).

b. Wacław Sierpiński, "Remarque sur la théorème de Riemann relatif aux séries semi-convergentes", Prac. Mat. Fiz. XXI (1910) 17–20 (in Polish).

c. Wacław Sierpiński, "Sur une propriété des séries qui ne sont pas absolument convergentes", Bull. Intern. Acad. Sci.: Cracovie A (1911) 149–158.

Sierpiński proved three extensions of Riemann's theorem:

Let $\sum_{n\ge0}a_n=s$ be a conditionally convergent series. Then:

  1. For every $r\in{\mathbb R}\cup\{-\infty,+\infty\}$ there is a permutation $\pi:{\mathbb N}\to{\mathbb N}$ such that $\sum_{n=0}^\infty a_{\pi(n)}=r$ and for every $n$, $$ a_n<0\Longleftrightarrow a_{\pi(n)}<0. $$
  2. For every $r\le s$ there is a permutation $\pi$ with $\sum a_{\pi(n)}=r$ and $\pi(n)=n$ whenever $a_n<0$.
  3. Fore very $r\ge s$ there is a permutation $\pi$ such that $\sum a_{\pi(n)}=r$ and $\pi(n)=n$ whenever $a_n>0$.

The result asked in the original question is item 3 above.

There is a recent paper that addresses similar questions in a more general (descriptive set-theoretic) context, and I believe you may find interesting some of the results there. The paper is "Rearrangement of conditionally convergent series on a small set" by Rafał Filipów and Piotr Szuca, Journal of Mathematical Analysis and Applications, 362 (2010) 64–71.

In this paper, Riemann's theorem is extended in a spirit similar to Sierpiński's results: Now one considers an ideal ${\mathcal I}\subseteq{\mathcal P}({\mathbb N})$, and asks whether for every conditionally convergent series as above and any $r$ in the extended reals, there is a permutation $\pi$ of the natural numbers such that $\sum a_{\pi(n)}=r$ and $$\{n\mid\pi(n)\ne n\}\in{\mathcal I}.$$ If this is the case, one says that ${\mathcal I}$ has property (R).

(The standard intuition behind an ideal is that it defines a notion of smallness, so we are asking that only a small number of indices are changed.) The results in the paper relate ideals with property (R) to other well-known classes of ideals, studied by Farah and others. Section 4 of the paper specifically deals with Sierpiński-like results for ideals with property (R), see their Theorem 4.1.

Here is a sketch of what I believe is a the proof of the result you are asking about. (I apologize if I am missing something obvious, let's hope that's not the case.)

Let $A=\{n\mid a_n\ge 0\}$, $B=\{n\mid a_n<0\}$. Since the series converges, $a_n\to0$. Since the convergence is conditional, both $\sum_{n\in A}a_n$ and $\sum_{n\in B}a_n$ diverge. It follows that we can find an infinite subset $C$ of $B$ such that $\sum_{n\in C}a_n$ converges.

I describe now how to rearrange the series by permuting the indices via a bijection $\pi$, so that it satisfies your requirement that $\pi(n)=n$ if $a_n\ge0$ and the new series converges to $s'$ (following your notation, $s'\gt s$ where $s$ is the value the original series converges to). I am assuming $s'$ is finite. A small modification of the argument gives the result for $s'=+\infty$.

To ease the presentation, my description is given by stages, as some kind of iterative process.

a. Find an $n_1$ as small as possible so that $\sum_{k\le n_1}a_{\pi(k)}\ge s'$, where we require that $\pi$ is injective on its domain, $\pi(k)=k$ if $a_k\ge 0$, and whenever $a_k<0$, we replace $a_k$ with $a_t$ where $t$ is least in $C$ that we have not used so far.

Note that we can find such $n_1$, because $\sum_{k\in A}a_k$ diverges but $\sum_{k\in C}a_k$ converges.

b. Find $n_2>n_1$ as small as possible so that $\sum_{k\le n_2}a_{\pi(k)}\le s'$, where we extend the previously defined partial $\pi$ so that still $\pi(k)=k$ if $a_k$ is positive, but now each $a_k<0$ is replaced with some $a_t$ with $t\in B$, so that the numbers in $B$ are used in increasing order, skipping only those that were already used as elements of $C$ in the previous stage. Also, $n_2$ should be large enough that at least one element of $B$ not previously considered is used.

The point is that $n_2$ exists, because otherwise this process simply gives us a rearrangement of the original series where only finitely many indices where switched, and so the sum converges to $s\lt s'$, but then the partial sums will all be smaller than $s'$ from some point on, contradicting the assumption that $n_2$ does not exist.

c. Find $n_3>n_2$ as small as possible so that $\sum_{k\le n_3}a_{\pi(k)}\ge s'$, and we extend the previously defined $\pi$ so now again only elements from $C$ not previously considered are used. Again, make sure at least one additional member of $A$ is added to the domain of $\pi$.

Again, $n_3$ exists for the same reason as in part a.

d. Find $n_4>n_3$ where now the sum is below $s'$ and we go through the elements of $B$ that have not been used so far in order, as in part b. Again, the argument in b. shows that $n_4$ exists.


The proof that this process works is very similar to the argument for the usual Riemann theorem.

(For the case $s'=+\infty$, in stage a. ensure that the sum is at least $s+1$, in b. that at least one new element of $B$ is used, in c. that it is at least $s+2$, in d. that at least one new element of $B$ is used, etc.)

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Recently, I've found mentioned Sierpinski's paper "Sur une propriete des series..." which contains the proof. It is in the 1st volume of selected papers by Sierpinski, the link to the whole volume is www.plouffe.fr/simon/math/Sierpinski%20Oeuvres%20Choisies%20I.pdf Original construction is different from the proposed one (I didn't succeed to check that Andreas' construction leads to the result) –  Georgii May 18 at 18:02

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