# Unconditional convergence

1. Oct 10, 2009

### 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 $$A=\sum_{i\in \mathbb{N}}\sum_{j\in \mathbb{N}}{a_{i,j}}$$??

Last edited: Oct 10, 2009
2. Aug 24, 2012

### 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