I'm told that the following is true when $m,n\in\Bbb Z$: $$\sum\limits_{i=1}^{m}\sum\limits_{j=1}^{n} \alpha_{ij}=\sum\limits_{j=1}^{n}\sum\limits_{i=1}^{m} \alpha _{ij}$$ When $m,n\to\infty$,
For any finite integers $m,n$, the expression $$ \sum_{i = 1}^m\sum_{j = 1}^n a_{ij} = \sum_{j = 1}^n \sum_{i = 1}^m a_{ij} $$ is true.
Your question should be interpreted as asking whether the "exchanging of limits" is allowed. I.e.
$$ \lim_{m\to\infty} \lim_{n\to\infty} \sum_{i = 1}^m\sum_{j = 1}^n a_{ij} \overset{?}{=} \lim_{n\to\infty}\lim_{m\to\infty} \sum_{j = 1}^n \sum_{i = 1}^m a_{ij} \tag{*}$$
Note that the left and right hand sides are often written, respectively, as $\sum_{i = 1}^\infty\sum_{j = 1}^\infty a_{ij}$ and $\sum_{j = 1}^\infty \sum_{i = 1}^\infty a_{ij}$.
(*) is true when $a_{ij}$ is absolutely summable. But in general there can be counterexamples. A particular one that comes to mind is
$$ a_{ij} = \begin{cases} 1 & i = j \\ -1 & i = j-1 \\ 0 & \text{otherwise}\end{cases} $$
Observe that $\sum_{j = 1}^\infty a_{ij} = 1 + (-1) = 0$ for any $i$. So the LHS of (*) evaluates to 0. On the other hand,
$$\sum_{i = 1}^\infty a_{ij} = \begin{cases} 1 & j = 1 \\ 0 & j > 1\end{cases} $$ so that the RHS of (*) evaluates to 1.
Now, for a more spectacular example, let $$ a_{ij} = \begin{cases} i & i = j \\ -i & i = j-1 \\ 0 & \text{otherwise}\end{cases}$$ The LHS still evaluates to 0. For the RHS however, notice that $\sum_{i = 1}^\infty a_{ij} = 1$ for every $j$, and so the RHS does not even converge!
