Let $f_m:\mathbb N\to \mathbb R^+\cup\{+\infty \}$ an increasing sequence. Set $$f(n):=\lim_{m\to \infty }f_m(n)$$
(it can possibly be $\infty $).
I want to prove that $$\lim_{m\to \infty }\sum_{n\in\mathbb N} f_m(n)=\sum_{n\in\mathbb N}f(n).$$ First, $$\sum_{n\in\mathbb N}f_m(n)\leq \sum_{n\in\mathbb N}f(n)$$ is clear, and thus $$\lim_{m\to \infty }\sum_{n\in\mathbb N}f_m(n)\leq \sum_{n\in\mathbb N}f(n).$$
I have problem for the reverse inequality. I tried to use the fact that $f(n)=\sup_{m\in\mathbb N}f_m(n).$ So if $\varepsilon>0$, there is $M_n$ s.t. $$f(n)-\varepsilon\leq f_{m}(n)\leq f(n),$$ for all $m\geq M_n$. Now since $\sum_{n\in\mathbb N}\varepsilon$ doesn't converge, I can't conclude.
Context
On $(\mathbb N,\mathbb P(\mathbb N),\mu)$ where $\mu(A)=\# A$, I want to show that $$\int_{\mathbb N}f(n)d\mu(n)=\sum_{n\in\mathbb N}f(n).$$
If $f$ is simple it work. Now let $f$ measurable. There is a sequence of simple function s.t. $f_m(n)\nearrow f(n)$. Then, using Monotone convergence, $$\int_{\mathbb N} f(n)d\mu(n)=\lim_{m\to \infty }\int_{\mathbb N}f_m(n)d\mu(n)=\lim_{m\to \infty }\sum_{n\in\mathbb N}f_m(n).$$ So now, I have to prove that $$\lim_{m\to \infty }\sum_{n\in\mathbb N}f_m(n)=\sum_{n\in\mathbb N}f(n),$$ to conclude.
First, the last section seems a little silly. If you're willing to use measure theory then there's nothing to prove; MCT states precisely that if $f_{m+1}\ge f_m\ge0$ then $\lim\int f_m=\int\lim_m f_m$, qed. (Hint for showing that if $\mu$ is counting measure and $f\ge0$ then $\int_{\Bbb N}f\,d\mu=\sum_nf(n)$: By definition $\sum_nf(n)=\lim_{N\to\infty}\sum_{n=1}^Nf(n)$.)
There's a very simple elementary proof. Let $$\sum_n f(n)=s\in[0,\infty].$$ Monotonicity makes it clear that the limit exists and that $$\lim_m\sum_nf_m(n)\le s.$$
For the other direction, suppose that $\alpha<s$. (Note I'm talking about a general $\alpha<s$ instead of talking about $s-\epsilon$ in order to include the possibility that $s=\infty$.) Choose $N$ so that $$\sum_{n=1}^Nf(n)>\alpha.$$ Then it's clear that there exists $M$ so $$\sum_{n=1}^Nf_m(n)>\alpha\quad(m>M).$$Hence $$\lim_m\sum_nf_m(n)>\alpha,$$and since this holds for every $\alpha<s$ we must have $$\lim_m\sum_nf_m(n)\ge s.$$