Changing limits in absolutely convergent series

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.

Solutions Collecting From Web of "Changing limits in absolutely convergent series"

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 $f(n,m) \le g(n) \quad\forall m\in\mathbb N$, 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

$$\sum_{n=0}^\infty \lim_{m\to\infty} f(n,m) = \sum_{n=0}^\infty 0 = 0$$