Under what condition we can interchange order of a limit and a summation? Suppose f(m,n) is a double sequence in $\mathbb R$. Under what condition do we have $\lim\limits_{n\to\infty}\sum\limits_{m=1}^\infty f(m,n)=\sum\limits_{m=1}^\infty \lim\limits_{n\to\infty} f(m,n)$? Thanks!
 A: 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. 
A: 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 $f(m,n)=\sum_{j=1}^n a_{m,j}$.
A: In this answer, I will focus on uniform convergence. That has been discussed in the comments, but not in any proper answer.
The interchange is valid if the partial sums are uniformly convergent, in the sense that $$\sup_{n}\left|\sum_{m>N} f(m,n)  \right| \to 0 \ \  \text{as} \ N \to \infty  \ \ \ \ \ \ \ \  \ \ (1)$$
More precisely, we show that if (i) $\sum_{m} f(m,n)$ converges for each $n$, (ii) $\lim_{n} f(m,n)$ exists for each $m$ and (iii) $(1)$ holds, then $\sum_{m} \lim_{n} f(m,n)$, $\lim_{n} \sum_{m} f(m,n)$ both exist and are equal.
There are cases where the above theorem can be applied, but neither the DCT nor the MCT can. An example is something like $f(m,n) = 2^{-(mn+1)} + \frac{(-1)^m}{m}$. It is a weaker condition than that required by the dominated convergence theorem, in the sense that if the DCT applies to $f(m,n)$ i.e., $|f(m,n)| \leq K_{m}$, where $\sum_{m} K_{m} < \infty$ then $(1)$ holds for $g$, by the Weierstrass M-test.
The proof follows from following fact: if $g_n \to g$ uniformly in a metric space $E$, with the $g_n$'s continuous, then $g$ is continuous. With a bit of work, this theorem can be applied in our context. The idea is to identify convergent sequences $(a_n)_n$ as precisely the continuous functions on $E=\{1/n\} \cup \{0\}$ with the induced (Euclidean) metric. If we are convinced of this identification, let us define $g_N(1/n) = \sum_{m=1}^{N} f(m,n)$. This is a finite sum, hence in light of (ii), we deduce $$\lim_{1/n \to 0^{+}} g_{N}(1/n) = \sum_{m=1}^{N} \lim_{n} f(m,n)$$ Thus defining $g_{N}(0) =   \sum_{m=1}^{N} \lim_{n} f(m,n)$ we have that the $g_{N}$'s are continuous on $E$ and by $(1)$ converge uniformly on $E \setminus \{0\}$. It is not hard to check that this implies they converge uniformly on $E$. Indeed, if $|g_{N'}(x) - g_{N''}(x)| < \epsilon$ for $N', N''>M_{\epsilon}, x \in E \setminus \{0\}$ we can take $x \to 0^{+}$ to get $|g_{N'}(0) - g_{N''}(0)| \leq \epsilon$ since the $g_{N}$'s are continuous. Hence $$g(0) = \lim_{N} g_{N}(0) = \sum_{m} \lim_{n} f(m,n)$$ is well-defined, and since $g$ is continuous (as the uniform limit of continuous functions), we deduce $$\sum_{m} \lim_{n} f(m,n) = g(0)= \lim_{1/n \to 0^{+}} g(1/n) = \lim_{n} \sum_{m} f(m,n) $$

However, it is very important to be careful. The result above only applies if the partial sums converge uniformly. It is possible to have functions $f_n:\mathbb{N} \to \mathbb{R}$ such that $f_n \to f$ uniformly i.e., $\sup_{m} |f_n(m) - f(m)| \to 0$ but $\lim_{n} \sum_{m} f_n(m) \neq  \sum_{m} \lim_{n} f_n(m)$. For instance, define $f_n(m)=1/n$ for $1 \leq m \leq n$, $f_n(m)=0$ for $m>n$. Then we can check that $\sup_{m} |f_n(m)| \leq 1/n \to 0$ but $1 = \lim_{n} \sum_{m} f_n(m) \neq  \sum_{m} \lim_{n} f_n(m) = 0$.

A measure-theoretic generalization of these ideas is that of uniform integrability.
Uniform integrability can be used, among other things, to a prove stronger version of the DCT, the Vitali convergence theorem.
A: 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:
$$ \lim_{n \rightarrow \infty} \sum_{m=1}^\infty f(m,n) \geq \sum_{m=1}^\infty \left( \lim_{n \rightarrow \infty} f(m,n) \right). $$
A: Denote $h(m)=\lim_{n\to\infty}f(m,n)$, and suppose that $\sum_{m=1}^\infty h(m)<\infty$. Then my guess is that the sufficient condition is
\begin{align}
\limsup_{n,m}\left|\sum_{k=1}^mf(k,n)-\sum_{k=1}^m h(k)\right|=0.
\end{align}
