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Math Help - Rearrangement of conditionally convergent series

  1. #1
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    Rearrangement of conditionally convergent series

    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
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  2. #2
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    Quote Originally Posted by Georgii View Post
    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
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  3. #3
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    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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