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 $s'+\varepsilon?$


Solution 1:

This question was already answered at MO: https://mathoverflow.net/questions/47589/

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