Does Unconditional Convergence Apply to Infinite Subseries Partitions?

[micromass]

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

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

I know this holds if every A_n 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 \sum_{(i,j)\in \mathbb{N}\times\mathbb{N}}{a_{i,j}}. This series is unconditional convergent in the following sense: for every bijection f:\mathbb{N}\rightarrow\mathbb{N}\times\mathbb{N} the series \sum_{n\in \mathbb{N}}{a_{f(n)}} is convergent to the same limit, say A. Now does A=\sum_{i\in \mathbb{N}}\sum_{j\in \mathbb{N}}{a_{i,j}}??

 

[micromass]

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

\sum_n x_n = \sum_n \sum_{m\in A_n} x_m

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

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

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

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

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

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

\|x-\sum_{n\leq k}\sum_{m\in B_n} x_m\|\leq \varepsilon

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