Under what condition we can interchange order of a limit and a summation?
A fairly general set of conditions, sufficient for many applications, is given by the hypotheses of dominated convergence. (Note that sums are just integrals with respect to the counting measure on $\mathbb{N}$, so dominated convergence applies with no modification.)
Without domination, the idea is that lumps of positive mass can "escape to infinity" when one attempts to interchange sum and limit. Here is a basic example: let $f_{m,n} = 1$ if $m = n$ and $0$ otherwise. Then $\sum_{m=1}^{\infty} f(m, n) = 1$ for all $n$, so the LHS is $1$, but $\lim_{n \to \infty} f(m, n) = 0$, so the RHS is $0$. The point of domination is to prevent these lumps of mass from escaping.
Not an answer to your question as such, but a note which seems worth making. Again from the point of view of measure theory (as previously mentioned by Qiaochu Yuan), you can use Fatou's Lemma to show that you have:
$$ \lim_{n \rightarrow \infty} \sum_{m=1}^\infty f(m,n) \geq \sum_{m=1}^\infty \left( \lim_{n \rightarrow \infty} f(m,n) \right). $$
Another sufficient condition is monotone convergence, which is the condition that for each $m$ the sequence $(f(m,n))$ is non-decreasing. The result can be proved via the machinery of measure theory; what follows is an elementary proof.
Claim: Assume $\sum_m f(m,1)>-\infty$. If for each $m$ we have $f(m,n)\le f(m,n')$ whenever $n\le n'$, then $$\lim_n\sum_{m=1}^\infty f(m,n) = \sum_{m=1}^\infty\lim_n f(m,n).\tag1$$
Proof: Wlog $f(m,n)\ge0$ for all $m,n$; else replace $f(m,n)$ with $f(m,n)-f(m,1)$. Define for each $k,n$ the partial sum $S_{k,n}:=\sum_{i=1}^k f(i,n)$. Write $S_k:=\lim_n S_{k,n}=\sum_{i=1}^k\lim_n f(i,n)$, and $S:=\sup_{(k,n)} \{S_{k,n}\}$. Note that monotonicity implies that $S_k$ exists, possibly with value infinity. Argue that: (a) the sequence $(S_k)$ is nondecreasing; (b) $S_k\ge S_{k,n}$ for each $k, n$; and (c) $S\ge S_k$ for all $k$. These three facts imply that $$ \lim_k S_k = S,\tag2 $$ regardless of whether the supremum $S$ is finite or infinite. Next, define analogously $T_{k,n}:=\sum_{i=1}^n f(i,k)$, let $T_k:=\lim_n T_{k,n}$, and $T:=\sup_{(k,n)} \{T_{k,n}\}$. Again argue that the sequence $(T_k)$ is nondecreasing; that $T_k\ge T_{k,n}$ for each $k,n$; and that $T\ge T_k$ for all $k$; and deduce that $$ \lim_k T_k=T.\tag3 $$ The proof is complete after observing that $$ \lim_k S_k = \lim_{k\to\infty}\sum_{i=1}^k \lim_n f(i,n)=\text{RHS of (1)},$$ and $$ \lim_k T_k=\lim_{k\to\infty}\sum_{i=1}^\infty f(i,k)=\text{LHS of (1)}, $$ and finally that $S$ and $T$ are the same thing, since $T_{k,n}=S_{n,k}$.
The above claim justifies interchanging the order of summation for a double series of non-negative terms:
Corollary: If $a_{i,j}\ge0$ for all $i$ and $j$, then $$\sum_{m=1}^\infty\left(\sum_{n=1}^\infty a_{m,n}\right) =\sum_{n=1}^\infty\left(\sum_{m=1}^\infty a_{m,n}\right).$$
Proof: Apply the claim with $f(m,n)=\sum_{j=1}^n a_{m,j}$.