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## How to rearrange only negative part of a conditionally convergent series to get any sum greater then initial?

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 2010 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 2010 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 2010 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 2010 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 2010 at 2:31
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Georgii:

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.

Etc.

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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