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Consider:$$\sum_{m,n \in \mathbb{N}_0} \frac{(-m)^n}{n!}.$$

I noticed that $$\sum_{m \in \mathbb{N}_0} \left(\sum_{n \in \mathbb{N}_0} \frac{(-m)^n}{n!} \right)= \sum_{m \in \mathbb{N}_0} e^{-m} =\frac {e} {e-1} . $$ but since the series $\sum_{m,n \in \mathbb{N}_0} \frac{(-m)^n}{n!}$ is not absolutely convergent I can't conclude the series converges. Any help would be greatly appreciated.

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    What definition of convergence of a double series do you use? – 2017-01-03
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    By convergence I mean, $\sum_m \sum_n x_{m,n}=\sum_n \sum_m x_{m,n}$ – 2017-01-03
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    Then no. Only the nesting with the inner sum over $n$ is convergent, with the inner sum over $m$, the inner series diverges for every $n$. – 2017-01-03

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