Let \(E\) be the set \(\{ \nicefrac{1}{n} : n \in \N \} \cup \{ 0 \}\text{,}\) and treat it as a metric space with the metric inherited from \(\R\text{.}\) Define the sequence of functions \(f_k \colon E \to \C\) by

\begin{equation*}
f_k(\nicefrac{1}{n}) \coloneqq \sum_{m=1}^n a_{k,m}
\qquad
\text{and}
\qquad
f_k(0) \coloneqq \sum_{m=1}^\infty a_{k,m} .
\end{equation*}

As the series converges, each \(f_k\) is continuous at \(0\) (since 0 is the only cluster point, they are continuous at every point of \(E\text{,}\) but we don’t need that). For all \(x \in E\text{,}\) we have

\begin{equation*}
\sabs{f_k(x)} \leq \sum_{m=1}^\infty \sabs{a_{k,m}} .
\end{equation*}

As \(\sum_k \sum_m \sabs{a_{k,m}}\) converges (and does not depend on \(x\)), we know that

\begin{equation*}
\sum_{k=1}^n f_k(x)
\end{equation*}

converges uniformly on \(E\text{.}\) Define

\begin{equation*}
g(x) \coloneqq \sum_{k=1}^\infty f_k(x) ,
\end{equation*}

which is, therefore, a continuous function at \(0\text{.}\) So

\begin{equation*}
\begin{split}
\sum_{k=1}^\infty \left( \sum_{m=1}^\infty a_{k,m} \right)
& =
\sum_{k=1}^\infty f_k(0)
= g(0)
= \lim_{n\to\infty} g(\nicefrac{1}{n}) \\
&=
\lim_{n\to\infty}\sum_{k=1}^\infty f_k(\nicefrac{1}{n})
=
\lim_{n\to\infty}\sum_{k=1}^\infty \sum_{m=1}^n a_{k,m} \\
&=
\lim_{n\to\infty}\sum_{m=1}^n \sum_{k=1}^\infty a_{k,m}
=
\sum_{m=1}^\infty \left( \sum_{k=1}^\infty a_{k,m} \right) . \qedhere
\end{split}
\end{equation*}