Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Suppose that series $\sum^\infty_{n=1} u_n=s$ converges conditionally. Then for each $s'\gt s$ there exists a permutation of positive integers $\sigma:\mathbb{N}\to\mathbb{N}$ such that

  1. if $u_n\geq 0,$ then $\sigma(n)=n;$

  2. $\sum^\infty_{n=1} u_{\sigma(n)}=s'.$

The standard proof of Riemann's theorem fails because of the condition (1). I know one construction which may give needed permutation, but I don't know how to prove the convergence. The construction is as follows.

Denote $I_+=\{n\in \mathbb{N}:u_n\geq 0\}, \ I_-=\{n\in \mathbb{N}:u_n\lt 0\}$ - positions of positive and negative terms respectively. Choose infinite set $F\subset I_-,$ such that $\sum\limits_{n\in F} u_n\gt-\infty.$ The permutation $\sigma$ is built inductively. For all $n\in I_+$ define $\sigma(n)=n.$ Next, on every position $n\in I_-$ permutation $\sigma$ puts elements of $F$ in their order. There exists first number $n_1$ such that $\sum\limits^{n_1}_{n=1} u_{\sigma(n)}\gt s'.$ The existence follows from the divergence $\sum\limits_{n\in I_+} u_n=\infty$ and convergence $\sum\limits_{n\in F} u_n\gt-\infty.$ After $n_1$ on each $n\in I_-$ permutation $\sigma$ puts elements of $I_-$ which were not used before in their order. There exists first number $n_2$ such that $\sum\limits^{n_2}_{n=1} u_{\sigma(n)}\lt s'.$ Existence follows from the fact that from some moment we'll obtain the permutation of series, which changes only finite number of points and converges to $s\lt s'.$ After $n_2$ at each position $n\in I_-$ put elements of $F,$ which were not used before, until first moment $n_3$ for which $\sum\limits^{n_3}_{n=1} u_{\sigma(n)}\gt s'.$ And so on.

I know how to prove convergence of partial sums of new series to $s'$ for indices $n\in[n_{2p},n_{2p+1}],$ (where sums less then $s'$). Is it possible that infinitely often partial sums will exceed some $s'+\varepsilon?$

share|cite|improve this question
I believe this result is a theorem of Sierpinski. The proof is really not easy, but if I remember correctly it appears in at least one book that he wrote. – Pete L. Clark Jan 20 '12 at 4:52
It seems that this question was answered at MO: – Martin Sleziak May 22 '12 at 8:32
By the way, can someone point some book where I can find proof of version of Riemann theorem for infinite products? – data May 22 '12 at 10:06
@m.woj Probably the best think to do would be ask this as a separate question on the main site. (Of course, in the question you should explain what precisely you mean by Riemann theorem for products.) – Martin Sleziak May 22 '12 at 15:47

This question was already answered at MO:

I am posting this as a CW answer to avoid bumps by Community.

share|cite|improve this answer

The proof I found for this exercise is quite long and awkward and I do not know a simpler or elegant one. There is not enough space here to describe it in details. Nevertheless, my approach went this way : by recurrence, the principle is to move the big negative terms further towards infinity and replace them by smaller ones, and then the sum approaches and eventually goes strictly above $s'$. As soon as it happens begin using smaller negative terms till it goes below $s'$ again. Then you have to show that this oscillating behavior is actually converging towards $s'$ with smaller and smaller oscillation which is not very difficult AND that you actually have bijection at the end with no negative terms pushed indefinitely. The control of the "replacement" phase has to be careful for the recurrence to work and actually to have a bijection in the end. The way I did it was each time I wanted the sum to increase I put the negative terms away in a buffer one by one and in order of the index. Then as soon as I am above $s'$, I replace the negative terms by the ones in the buffer, FIFO, if any. In the end, because the series is a convergent one, I am sure that the oscillations will decrease and the buffer will be used up.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.