# 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 '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

Georgii:

Let me start with 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. For every $r\ge s$ 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. Note that items 1 and 2 follow easily from this.

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. MR2557668 (2010i:40001).

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 the proof of item 2. [Edit, Nov 17, 2014: My original argument for item 3 started with conditionally convergent series $\sum_i u_i=r$ and, for each $s\ge r$, produced a rearrangement fixing the positive terms, and whose partial sums have liminf $s$, but to argue that their liminf is also $s$ requires additional work. Fixing this lead me to a proof very similar to Sierpiński's, so I decided to reorganize the answer and just present his argument, together with what I hope are some additional explanations. What follows is part of a post on rearrangements of series on my blog.]

Theorem (Sierpiński). Let $\sum_i u_i$ be a conditionally convergent series of real numbers that converges to $U$. For any $V\le U$ there is a rearrangement $v_0,v_1,\dots$ of the $u_i$ such that $\sum_i v_i=V$, and the rearrangement leaves fixed all non-positive terms.

Similarly, for any $W\ge U$ there is a rearrangement $w_0,w_1,\dots$ of the $w_i$ such that $\sum_i w_i=W$, and the rearrangement leaves fixed all positive terms.

Moreover, for any extended real $\alpha$, there is a rearrangement $s_0,s_1,\dots$ of the $u_i$ such that $\sum_i s_i=\alpha$ and, for every $n$, $u_n>0$ iff $s_n>0$.

Proof. Note first that it suffices to prove the result about rearrangements fixing non-positive terms, since the other results are consequence of it: simply apply the first result to the series $\sum_i -u_i$, noting that $-W\le -U$, to obtain the second result. For the first, apply first either the first or the second result if $\alpha\in\mathbb R$. The case where $\alpha=\pm\infty$ is handled easily.

Just as Riemann's theorem depended on a lemma about the series of positive and non-positive terms of the original sequence, Sierpiński's result depends on a lemma about the series of positive terms. Unlike Riemann's theorem, in this case, all the work goes into the proof of the lemma, and the theorem is more properly a corollary of it.

Lemma. Suppose $a_0,a_1,\dots$ is a sequence of positive numbers such that $\sum_i a_i=+\infty$ and $\lim_i a_i=0$ Let $A_0,A_1,\dots$ denote the sequence of partial sums. For any $L\ge0$ there is a rearrangement $c_0,c_1,\dots$ with sequence of partial sums $C_0,C_1,\dots$ such that $\lim_i (A_i-C_i)=L$.

To see that the lemma implies the result, suppose that $V\le U$, and let $L=U-V$. Let $a_0,a_1,\dots$ and $-b_0,-b_1,\dots$ be, respectively, the subsequences of positive and of non-positive terms of the $u_i$. Let $A_0,A_1,\dots$ and $B_0,B_1,\dots$ denote, respectively, the sequences of partial sums of the $a_i$ and of the $b_i$. For each $i$, let $p_i$ be the number of positive terms among $u_0,\dots,u_i$, and let $n_i$ be the number of non-positive terms, so that $p_i+n_i=i+1$ and, if $U_0,U_1,\dots$ denote the partial sums of the $u_i$, then $U_i=A_{p_i}-B_{n_i}$ for all $i$, and $p_i,n_i\to+\infty$.

Apply the lemma to the $a_i$ to obtain a rearrangement $c_0,c_1,\dots$ with partial sums $C_0,C_1,\dots$ such that $A_i-C_i\to L$. Consider the rearrangement $v_0,v_1,\dots$ of the $u_i$ that fixes the $-b_i$ and permutes the $a_i$ as indicated. If $V_0,V_1,\dots$ are the partial sums of the $V_i$, note that, by design, $V_i=C_{p_i}-B_{n_i}$, so that $U_i-V_i=A_{p_i}-C_{p_i}\to L$, and $V_i\to U-L=V$, as wanted. All that remains is to prove the lemma.

Proof. As in Riemann's theorem, we proceed by stages. Starting with $A_{-1}=C_{-1}=0$, at stage $n$ we examine $A_{n-1}-C_{n-1}$ to decide the value of $c_n$. As in that theorem, we want to arrange that the values $A_i-C_i$ increase if smaller than $L$, and decrease if larger, so that in the limit we obtain the desired value. Actually, we will need to modify slightly this strategy. To motivate the proof, consider first the case where the $a_i$ are monotonic, that is, $a_0\ge a_1\ge\dots$ In this case, the strategy works: At stage $n$, we consider two cases, according to whether $A_{n-1}-C_{n-1}\le L$ or $A_{n-1}-C_{n-1}>L$:

• If $A_{n-1}-C_{n-1}\le L$, then let $r$ be such that $a_r<a_n/2$, and set $c_n=a_r$. This is possible, since $a_r\to 0$.

Note that, for as long as $A_k-C_k\le L$, we stay in this case. Thus, if from some point $n$ on we are always in case 1, then for $k$ large enough we have $$A_k-C_k=(A_{n-1}-C_{n-1})+\sum_{i=n}^k (a_i-c_i)\ge(A_{n-1}-C_{n-1})+ \frac12\sum_{i=n}^k a_i\to\infty.$$

This is a contradiction, and indicates that repeated applications of this case always terminate, and lead to case 2. In particular, case 2 is considered infinitely often. Moreover, if $k>n-1$ is least such that $A_k-C_k>L$, then $A_k-C_k=(A_{k-1}-C_{k-1})+(a_k-c_k)\le L+a_k$.

• If $A_{n-1}-C_{n-1}>L$, then in particular $A_{n-1}\ne C_{n-1}$, so at least one of the $a_i$, $i<n-1$, has not been considered yet as value of a $c_j.$ Pick the least such index $i$, and define $c_n=a_i$.

Since this case applies infinitely often, the value of the least index $i$ such that $A_i$ is not one of the $c_j$ increases unboundedly. By design, no index $i$ is used more than once, and therefore this algorithm results in a rearrangement of the $a_i$. If we ever find an index $k$ such that (again) $A_k-C_k\le L$, note that $$A_k-C_k=(A_{k-1}-C_{k-1})+(a_k-c_k)>(A_{k-1}-C_{k-1})-c_k>L-c_k.$$

Note that our assumption that the $a_k$ are  decreasing ensures that, in case 2, (with notation as above) $A_n-C_n=(A_{n-1}-C_{n-1})+(a_n-c_n)\le$ $A_{n-1}-C_{n-1}$, since $a_i\ge a_n$. Moreover, it cannot be that we have equality from some point on, since the $a_k$ converge to $0$. This means that the values $A_k-C_k$ decrease.

Since $a_i,c_i\to 0$, it follows that, if case 1 is also applied infinitely often, then $A_i-C_i\to L$.  Therefore, to conclude, it suffices to argue that we cannot stay in case 2 forever. To see this, argue by contradiction, and assume that from $n-1$ on, we always stay in case 2. Let $j_m$ be the number of indices $i\le m$ such that $a_i$ is not one of the $c_k$, $k\le m$. For any $m\ge n$, since $A_{m-1}-C_{m-1}>L$, then $j_m\le j_{m-1}$, with equality if and only if $a_m$ is not one of the $c_k$, $k<m$. Since $C_{n-1}\ne A_{n-1}$, necessarily at least one of the $c_i$, $i\le n-1$, is $a_r$ for some $r\ge n$. If $r_0$ is the least such index $r$, then $j_{r_0}<j_{n-1}$, because $a_{r_0}$ is one of the $c_i$ with $i<r_0$, in fact $i<n$. Since we cannot have an infinite decreasing sequence of positive integers, this means that (we have a contradiction and) eventually we should be in case 1 again. This completes the proof in the case the $a_i$ are decreasing.

Let's now consider the general case. If we try to implement the argument we just described, we see that if we are ever in case 1, $A_k-C_k\le L$, the sequence of $A_i-C_i$ increases until we find an index $j$ with $A_j-C_j>L$. Also, once we enter case 2, we cannot stay there indefinitely, as the number of terms $a_i$ that are not some $c_k$, $k\le j$, decreases. The problem is that the sequence of $A_i-C_i$ is not necessarily decreasing. In fact, if we are at a stage where we define $c_n=a_i$ for some $i<n$, and it happens that $a_n>a_i$, then $A_n-C_n>A_{n-1}-C_{n-1}$. This means that, although we have ensured that $\liminf (A_i-C_i)=L$, and that there is a subsequence of the $A_i-C_i$ converging to $L$, it may well be the case that $\limsup (A_i-C_i)>L$.

Sierpiński deals with this situation in a clever fashion. He ensures that if we are in case 2, so that $c_n=a_i$ for some $i<n$, then also $a_n=c_k$ for some $k<n$, which means that $c_k$ was defined according to the prescription in case 1. He uses this to guarantee that $A_n-C_n$ cannot stray too far from $L$. In detail, Sierpiński proceeds as follows, defining now three cases. Assume we are at the beginning of stage $n$:

1. If $A_{n-1}-C_{n-1}\le L$, as before let $c_n$ be some $a_r$ where the index $r$ has not yet been chosen, and $a_r<a_n/2$, and $a_r<1/2^n$.
2. If $A_{n-1}-C_{n-1}>L$, now we consider two possibilities, according to whether or not the index $n$ was picked previously: If it was not, then we let $c_n=a_n$. Note that $A_n-C_n=A_{n-1}-C_{n-1}$ if this is the case, and that, since $C_{n-1}\ne A_{n-1}$, then necessarily some $c_i$ with $i<n$ must be $a_r$ for some $r>n$. This means that at some later stage (at most by stage $r$), we are no longer to be in this case.
3. If $A_{n-1}-C_{n-1}>L$, and the index $n$ was chosen previously, then we let $c_n$ be $a_i$, where $i$ is the first index less than $n$ not chosen yet.

As before, cases 1 and 3 happen infinitely often, so we have indeed defined a rearrangement. Moreover, if we are in case 3, then $a_n=c_i$ for some $i<n$ (this is why we included case 2). The point is that if stage $i$ is by case 2, then $c_i=a_i$, and if it is by case 3, then $c_i=a_j$ for some $j<i$. This means that the only way we can have $c_i=a_n$ for $n>i$ is if stage $i$ was by case 1, which means that $c_i<1/2^i$.

Fix $\epsilon>0$. Since cases 1 and 3 happen infinitely often, if we choose $n$ large enough, then we can ensure that all indices $m$ mentioned below are so large that $a_m,c_m<\epsilon$, and if $a_m=c_k$ for some $k<m$, then $k>N$ where $\sum_{i\ge N}1/2^i<\epsilon$. We want to ensure that $|(A_n-C_n)-L|<2\epsilon$. This proves that $A_n-C_n\to L$.

If stage $n$ is by case 1, and stage $m<n$ was largest where $A_m-C_m$ ${}>L$, then $A_{m+1}-C_{m+1}=(A_m-C_m)+(a_{m+1}-c_{m+1})>$ $L-c_{m+1}>L-\epsilon$. Since the sequence $A_i-C_i$ is increasing for $m+1\le i\le n$, we have $L\ge A_n-C_n>L-\epsilon$, as wanted.

Suppose then that stage $n$ is by case 3, and that stage $m<n$ was largest where we were in case 1. We then have that $A_{m+1}-C_{m+1}=(A_m-C_m)+(a_m-c_m)\le L+a_m<L+\epsilon$. Also, $$A_n-C_n=A_{m+1}-C_{m+1}+\sum_{i=m+2}^n(a_i-c_i),$$ but for any such $i$, we have that stage $i$ was either by case 2, and therefore $a_i-c_i=0$, or else it was by case 3, and therefore $a_i=c_j<1/2^j$ for some $j<i$, but necessarily $j>N$, so $$\sum_{i=m+2}^n(a_i-c_i)=\sum\{a_i-c_i\mid m+2\le i\le n,\mbox{ and stage }i\mbox{ was by case 3}\}\le\sum\{a_i\mid m+2\le i\le n,\mbox{ and stage }i\mbox{ was by case 3}\} \le\sum_{j>N}\frac1{2^j}<\epsilon.$$

This means that $A_n-C_n<L+2\epsilon$. But also $$A_n-C_n=(A_{n-1}-C_{n-1})+(a_n-c_n)>L-c_n>L-\epsilon.$$ This completes the proof. $\Box$

This concludes the proof of Sierpiński's theorem. $\Box$

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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
@Georgii I have updated the answer with a sketch of Sierpiński's approach. My argument was indeed incomplete. –  Andres Caicedo Nov 17 at 8:07