Interversion of limit and summation of double-indexed sequence Assume we have a sequence $(a_{m,n}) \in [0,1]^{\mathbb{N}\times \mathbb{N}}$ such that $a_{m,n} \rightarrow a_n\in [0,1]$ when $m \rightarrow \infty$. Are there any known conditions under which
$\sum_{n=0}^{\infty} a_{m,n} \rightarrow \sum_{n=0}^{\infty} a_n$ when $m \rightarrow \infty$ ? Assume it is known that both series converge.
 A: You can conclude that $\sum_{n=0}^{\infty} a_{m,n} \rightarrow \sum_{n=0}^{\infty} a_n$ if (but not only if) there exists $b_n \geq 0$ such that $\sum_0^{\infty}b_n < +\infty$ where $|a_{m,n}| \leq b_n$ for all $n,m \in \mathbb{N}$. This is a version of the famous dominated convergence theorem where the measure space is $(\mathbb{N},\mu)$ with $\mu(A)=\# A$.
A: We can translate Lebesgue's monotone and dominated convergence theorems to sequences $f_m = (a_{m, n})_{n=1}^\infty$ and $f = (a_n)_{n=1}^\infty$.
Lebesgue's monotone convergence theorem tells us that if $f_m$ is increasing then $$\int_\mathbb{N} f_m\ \mathrm{d}\mu \to \int_\mathbb{N} f\ \mathrm{d}\mu$$ where $\mu$ is the counting measure on $\mathbb{N}$. In other words, if $a_{m, n}$ is increasing in $m$ for each $n$, then $$\sum_{n=1}^\infty a_{m, n} \to \sum_{n=1}^\infty a_n.$$ But, clearly, we can just assume $a_{m, n}$ is increasing in $m$ for each big enough $n$ and $m$. And by that I mean, if there exist $N$ and $M$ such that $a_{m, n}$ is increasing in $m$ for $m>M$ and $n>N$.
Other theorem is the Lebesgue's dominated convergence theorem. It says that if there exists a function $g = (b_n)_{n=1}^\infty$ which bounds $f_m\leq g$ from above, meaning $a_{m, n}\leq b_n$ for all $m, n$, and $g$ is integrable, that is, $$\int_\mathbb{N} g\ \mathrm{d}\mu = \sum_{n=1}^\infty b_n < \infty,$$ then we can say that again $$\sum_{n=1}^\infty a_{m, n}\to \sum_{n=1}^\infty a_n.$$
Those are very important theorems in the theory of Lebesgue integrals. By treating infinite sums as integrals we can derive statements about infinite sums like above.
