Let $\sum_{n=0}^\infty f(n,m)$ be a real series. Suppose the series converges absolutely. Can we do the following? $$ \lim_{m\to\infty}\sum_{n=0}^\infty f(n,m)=\sum_{n=0}^\infty \lim_{m\to\infty}f(n,m)\quad $$
I thought about some series, but all of them fit. Is there any counterexample?
We suppose that all limits exist.
It looks like $f:\mathbb N^2 \to \mathbb R$. If not so, please correct me.
In this case it's indeed possible if $f(\cdot, m)$ is uniformly bounded by an absolutely summable sequence $g:\mathbb N\to\mathbb R$ by Lebesgue convergence theorem. Just view $f(\cdot, m)$ as a sequence of functions $f_m \in L^1(\mathbb N)$ where the space is equipped with the counting measure (also written as $\ell^1 (\mathbb N)$). If you have a sequence $g\in\ell^1(\mathbb N)$ such that $\left| f(n,m)\right| \le g(n) \quad\forall m\in\mathbb N$, then $$\lim_{m\to\infty} \sum_{n=1}^\infty f(n,m) = \sum_{n=1}^\infty \lim_{m\to\infty} f(n,m)$$ assuming the latter limits exist. However without this additional assumption, you can't relate the two limits.
If this does not hold, look at $f(n,m) = \delta_{nm}$ as a classical example: $\sum_{n=1}^\infty f(n,m) = 1$ and $\lim_{m\to\infty} f(n,m) = 0$ so $$\lim_{m\to\infty} \sum_{n=1}^\infty f(n,m) = 1\\ \sum_{n=1}^\infty \lim_{m\to\infty} f(n,m) = 0$$
Given your currenct assumptions, the answer is no.
For example, if $$f(n,m) = \begin{cases}1 & \text{if } m=n\\ 0&\text{else}\end{cases}$$ then the left limit is
$$\lim_{m\to\infty} \sum_{n=0}^{\infty} f(n,m)= \lim_{m\to\infty} 1 = 1$$
while the right is
$$\sum_{n=0}^\infty \lim_{m\to\infty} f(n,m) = \sum_{n=0}^\infty 0 = 0$$
