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 $n,m\in\mathbb{N}$  $$a_{nm}= \left\{ \begin{array}{rc}
1, &amp; n=m\\ -1, &amp; m=n+1 \\ 0, &amp; \text{in other case} \end{array} \right.$$

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...
Let $s_m:=\sum_{n=1}^{\infty}a_{nm}$ and observe that $s_1=1$ and $s_m=0$ for $m&gt;1$

That tells us that:  $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}a_{nm}=\sum_{m=1}^{\infty}s_m=1+0+0+0+\cdots=1$$

On base of that conclude that $$\sum_{m=1}^{\infty}\sum_{n=1}^{\infty}a_{nm}=1$$

Let $t_n:=\sum_{m=1}^{\infty}a_{nm}$ and observe that $t_n=0$ for every $n\in\mathbb N$.

That tells us that:  $$\sum_{n=1}^{\infty}\sum_{m=1}^{\infty}a_{nm}=\sum_{n=1}^{\infty}t_n=0+0+0+0+\cdots=0$$
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 $m&gt;1$, 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.