Calcule of $\sum_{m=1}^\infty\sum_{n=1}^\infty{a_{nm}}$ and $\sum_{n=1}^\infty\sum_{m=1}^\infty{a_{nm}}.$ Let for all $n,m\in\mathbb{N}$ $$a_{nm}= \left\\{ \begin{array}{rc} 1, & n=m\\\ -1, & m=n+1 \\\ 0, & \text{in other case} \end{array} \right.$$ Calcule: 1) $\,\,\,\,\displaystyle\sum_{m=1}^\infty\sum_{n=1}^\infty{a_{nm}}$. 2) $\,\,\,\,\displaystyle\sum_{n=1}^\infty\sum_{m=1}^\infty{a_{nm}}.$ Intiutively I think that both sums are zero, but I don't can't to write this decently...

Lets do part $1$ "formally":

If $m=1$, and study $n=1,2,\cdots$, the only non-zero value is at $n=1$ where $a_{nm}=1$.

For any other fixed $m>1$, if we run through the sum $\sum_{n=1}^\infty a_{nm}$, exactly two values of $n$ will have non-zero values, namely, $n=m-1$ and $n=m$ with values $-1$ and $1$ respectively so they cancel each other yielding a contribution of $0$.

So all these sums are $0$. Thus $$\sum_{m=1}^\infty \sum_{n=1}^\infty a_{nm}=\sum_{n=1}^\infty a_{n1} + \sum_{m=2}^\infty \sum_{n=1}^\infty a_{nm}=1+0=1$$

Now try doing the other part by yourself.