Does the limit of series of sequences equal series of limit of sequences? Let $ \{\{ a_n\}_m\} $ be a sequence of converging sequences, such that $ \sum_{m=0}^{\infty}\{a_n\}_m $ converges absolutely for all $ n $.
Is it true that $ \lim_n \sum_{m=0}^{\infty} \{a_n\}_m = \sum_{m=0}^{\infty}\lim_n \{a_n\}_m $?
 A: No (I understood the question differently from Gerry). Take $a_n(m) = 1/n$ if $m \lt n$ and $a_n(m) = 0$ otherwise (here $a_0(m) = 0$ is understood).
Then $\lim_m a_n(m) = 0$  for all $n$, hence each sequence $(a_n(m))_{m=0}^\infty$ converges.
On the other hand, $\sum_{m=0}^{\infty} a_n(m) = \sum_{m=0}^{n-1} \frac{1}{n} = 1$ for all $n \geq 1$ while $\lim_{n} a_n(m) = 0$ for all $m$ so that $\sum_{m=0}^\infty \lim_{n} a_n(m) = 0$.

Alternatively, take $a_{n}(m) = 1$ if $n=m$ and $a_{n}(m) = 0$ otherwise. Then the same properties as above hold.
A: Perhaps an even simpler counterexample would be
$$ a_{n,m} = \begin{cases}1 & \text{if } n = m \\ 0 & \text{otherwise.}\end{cases} $$
Then


*

*$\displaystyle \lim_{n\to\infty} a_{n,m} = 0$ for all $m$,

*$\displaystyle \sum_{m=0}^\infty |a_{n,m}| = \sum_{m=0}^\infty a_{n,m} = 1$ for all $n$,


and therefore


*

*$\displaystyle \lim_{n\to\infty} \sum_{m=0}^\infty a_{n,m} = 1$, but

*$\displaystyle \sum_{m=0}^\infty \lim_{n\to\infty} a_{n,m} = 0$.

A: The subscripts in the original are out of whack, so maybe I misunderstand, but if $a_n=1,0,0,0,0,\dots$ for even $m$, and $a_n=0,1,0,0,0,0,\dots$ for odd $m$, then the limit on the left is 1, but the sum on the right doesn't exist because $\lim_na_n$ doesn't exist. 
EDIT: now that the problem statement has been edited into something that makes sense, I'll piggyback on the answers that have been posted and note that if we take $a_{n,m}=n$ if $m=n$ and $0$ otherwise, then the limit of the sums is infinite, while the sum of the limits is zero, so they can be about as different as you like. Or take $a_{n,m}=(-1)^n$ if $m=n$, $0$ otherwise, and the limit of the sums doesn't exist, while the sum of the limits is $0$. 
