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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.

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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

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

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