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Unconditional convergence

  1. Oct 10, 2009 #1

    micromass

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    I have a unconditional serie [tex]\sum_n {x_n}[/tex] so for every permutation [tex]s:\mathbb{N}\rightarrow \mathbb{N}[/tex] the series [tex]\sum_n{x_{s_n}}[/tex] converges to the same limit.

    But let [tex]\{A_n\}_n[/tex] be a countable partition of N. Does the series [tex]\sum_n (\sum_{m\in A_n} {x_m})[/tex] converge (and if yes: to the same limit?)

    I know this holds if every [tex]A_n[/tex] is finite. But what if they are all infinite? Because then we have a series of infinite series and I can't find the permutation.



    Edit:
    The reason why I am asking: I have series [tex]\sum_{(i,j)\in \mathbb{N}\times\mathbb{N}}{a_{i,j}}.[/tex] This series is unconditional convergent in the following sense: for every bijection [tex]f:\mathbb{N}\rightarrow\mathbb{N}\times\mathbb{N}[/tex] the series [tex]\sum_{n\in \mathbb{N}}{a_{f(n)}}[/tex] is convergent to the same limit, say A. Now does [tex]A=\sum_{i\in \mathbb{N}}\sum_{j\in \mathbb{N}}{a_{i,j}}[/tex]??
     
    Last edited: Oct 10, 2009
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  3. Aug 24, 2012 #2

    micromass

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    I've been asked if I ever found the answer to this, and the answer is that I did. The answer is positive, that is: we can prove that

    [tex]\sum_n x_n = \sum_n \sum_{m\in A_n} x_m[/tex]

    Now, firstly, if a series converges unconditionally, then every subseries converges.

    Second, notice that unconditional convergence of [itex]\sum_n x_n[/itex] implies the following (actually equivalent) property:

    For every [itex]\varepsilon >0[/itex], there is a finite subset [itex]A\subseteq \mathbb{N}[/itex], such that for all finite subsets B with [itex]A\subseteq B\subseteq \mathbb{N}[/itex] we have that
    [tex]\|x-\sum_{n\in B} x_n\|\leq \varepsilon[/tex]

    Passing to the limit, we obtain the same inequality for infinite B.

    Now, consider a double series [itex]\sum_n \sum_{m\in B_n} x_m[/itex], where the [itex]B_n[/itex] are disjoint and conver [itex]\mathbb{N}[/itex].

    Apply the above criterion and obtain for given [itex]\varepsilon>0[/itex], the corresponding [itex]A\subseteq \mathbb{N}[/itex]. Since A is finite, we have [itex]A\subseteq \bigcup_{m\leq k}{B_m}[/itex]. So

    [tex]\|x-\sum_{n\leq k}\sum_{m\in B_n} x_m\|\leq \varepsilon[/tex]

    This is where I found the complete answer: http://www.artofproblemsolving.com/Forum/viewtopic.php?p=1676569
     
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