Double sum divergence

Question

I have been trying to prove $$ \sum_{m=1}^\infty\sum_{n=1}^\infty \frac{1}{(m+n)^2}=\infty $$

I tried interchanging the sums, but the problem essentially same. I am also aware that: For $a_{m,n}\in [0,\infty]$ $$ \sum_{m=1}^\infty\bigg(\sum_{n=1}^\infty a_{m,n}\bigg)=\sum_{k=1}^\infty a_{\psi(k)} $$ where $\psi:\mathbb{N}\to\mathbb{N}\times\mathbb{N}$ is a bijection. I could not find a way an appropriate $\psi$ to show that the latter one diverges. Any help is appreciated!

Answer 1

$$\sum_{m=1}^\infty\sum_{n=1}^\infty \frac{1}{(m+n)^2}\geq \sum_{m=1}^\infty\sum_{n=1}^m\frac{1}{(m+n)^2}\geq\sum_{m=1}^\infty\sum_{n=1}^m\frac{1}{(2m)^2}=\sum_{m=1}^\infty\frac{1}{4m}$$

Answer 2

$\begin{array}\\ s(N) &=\sum_{m=1}^N\sum_{n=1}^N \frac{1}{(m+n)^2}\\ &=\sum_{m=1}^N\sum_{n=m+1}^{m+N} \frac{1}{n^2}\\ &\gt\sum_{m=1}^N\sum_{n=m+1}^{m+N} \frac{1}{n^2}\\ &\ge\sum_{m=1}^N\sum_{n=m+1}^{m+N} \frac{1}{n(n-1)}\\ &\ge\sum_{m=1}^N\sum_{n=m+1}^{m+N}(\frac1{n-1}-\frac1{n})\\ &=\sum_{m=1}^N(\frac1{m}-\frac1{m+N})\\ &=\sum_{m=1}^N\frac1{m}-\sum_{m=1}^N\frac1{m+N}\\ &\gt\sum_{m=1}^N\frac1{m}-\sum_{m=1}^N\frac1{1+N}\\ &\gt\sum_{m=1}^N\frac1{m}-\frac{N}{1+N}\\ &\gt\sum_{m=1}^N\frac1{m}-1\\ \end{array} $

Since $\sum_{m=1}^N\frac1{m}$ diverges, so does $s(N)$.

Note that this only works with finite sums, so there are no problems with rearrangements or splitting a sum into two parts.