I want to show that the series $\displaystyle{\sum_{j,k}^{\infty}\frac{1}{j^2+k^2}}$ diverges.
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I have done the following:
We have that $(j+k)^2=j^2+2jk+k^2\geq j^2+k^2$.
We get that \begin{equation*}(j+k)^2\geq k^2+j^2 \Rightarrow \frac{1}{k^2+j^2}\geq \frac{1}{(j+k)^2}\end{equation*}
We rearrange the series and we get \begin{equation*}\sum_{j,k}^{\infty}\frac{1}{j^2+k^2}=\sum_{m=1}^{\infty}\sum_{j+k=m}\frac{1}{j^2+k^2}\end{equation*}
Then we have the following: \begin{align*}\sum_{m=1}^{\infty}\sum_{j+k=m}\frac{1}{j^2+k^2}&\geq \sum_{m=1}^{\infty}\sum_{j+k=m}\frac{1}{(j+k)^2}=\sum_{m=1}^{\infty}\sum_{j+k=m}\frac{1}{m^2}=\sum_{m=1}^{\infty}\frac{1}{m^2}\cdot \sum_{j+k=m}1 \\ & =\sum_{m=1}^{\infty}\frac{1}{m^2}\cdot (m-1)\end{align*} since there are $m-1$ terms in the sum $\sum_{j+k=m}$ (these are the pairs $(1, m-1), (2, m-2), \ldots (m-1, 1)$).
So, we have that \begin{equation*}\sum_{m=1}^{\infty}\sum_{j+k=m}\frac{1}{j^2+k^2}\geq \sum_{m=1}^{\infty}\frac{1}{m^2}\cdot (m-1)\end{equation*}
Is everything correct so far? How could we continue?
