Rearrangement of conditionally convergent series

**Georgii** · November 29th 2010, 12:02 PM

Suppose that $\sum^\infty_{n=1} u_n = s,$ where series converge conditionally and $s'>s.$ Help me please in proving following assertion: there exists permutation $\sigma:\mathbb{N}\rightarrow \mathbb{N}$ with properties:
1) $u_n\geq 0\implies \sigma(n)=n;$
2) $\sum^\infty_{n=1} u_{\sigma(n)}=s'.$

I can prove that for any sequences of positive numbers, $(b_n), (B_n)$ such that $b_n\rightarrow 0, \ \sum^\infty_{n=1} b_n=\infty, \ 0<B_1<B_2<\ldots<B_n\rightarrow \infty$ , there exists a permutation $\sigma:\mathbb{N}\rightarrow \mathbb{N}$ , for which
$\forall k\geq 1 \sum^k_{n=1} b_{\sigma(n)}\leq B_k.$
But I can't apply this properly to initial problem

**tonio** · November 29th 2010, 12:20 PM

Originally Posted by Georgii

Suppose that $\sum^\infty_{n=1} u_n = s,$ where series converge conditionally and $s'>s.$ Help me please in proving following assertion: there exists permutation $\sigma:\mathbb{N}\rightarrow \mathbb{N}$ with properties:
1) $u_n\geq 0\implies \sigma(n)=n;$
2) $\sum^\infty_{n=1} u_{\sigma(n)}=s'.$

I can prove that for any sequences of positive numbers, $(b_n), (B_n)$ such that $b_n\rightarrow 0, \ \sum^\infty_{n=1} b_n=\infty, \ 0<B_1<B_2<\ldots<B_n\rightarrow \infty$ , there exists a permutation $\sigma:\mathbb{N}\rightarrow \mathbb{N}$ , for which
$\forall k\geq 1 \sum^k_{n=1} b_{\sigma(n)}\leq B_k.$
But I can't apply this properly to initial problem

Read the proof of Riemann's Theorem in series.

Tonio

**Georgii** · November 29th 2010, 12:35 PM

My question deals nothing with the proof of Riemann's theorem, because the permutation which is built there doesn't satisfy property 1) of mentioned statement. And I don't know whether it is possible to build a permutation with properties 1) and 2) using the method of proof of Riemann's theorem.

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