How to prove $\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty$ How to prove $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=+\infty.$$
I try to do like $$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{n+m=N}^\infty  \frac{1}{m^2+n^2}=\sum_{N=1}^\infty \sum_{m=1}^{N-1}  \frac{1}{m^2+(N-m)^2}$$
 $$\frac{1}{m^2+(N-m)^2}\leq \frac{2}{N^2}$$
but it doesn't work.
 A: $$
\begin{align}
\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^2+n^2}
&\ge\sum_{m=1}^\infty\sum_{n=1}^\infty\frac1{m^2+n^2+2mn+m+n}\\
&=\sum_{m=1}^\infty\sum_{n=1}^\infty\left(\frac1{m+n}-\frac1{m+n+1}\right)\\
&=\sum_{m=1}^\infty\frac1{m+1}\\[6pt]
&=\infty
\end{align}
$$
A: If the sum were finite, then we could get a contradiction as follows. Breaking it up into 4 sums depending on whether or not $m$ and $n$ are even, we have 
$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}  = \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m)^2+(2n)^2} + \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m-1)^2+(2n)^2} $$
$$+ \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m)^2+(2n-1)^2} + \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m-1)^2+(2n-1)^2} $$
Note that each of the last three sums is greater than the first due to the denominators of each term being smaller. Thus we have
$$\sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{m^2+n^2}  > 4 \sum_{m=1}^\infty \sum_{n=1}^\infty \frac{1}{(2m)^2+(2n)^2} $$
But factoring out the 4 from the denominator we see the right hand side is the same as the left. Hence we get a contradiction and the sum is not finite.
A: You're almost there. 
$$\frac{1}{m^2 + (N-m)^2} = \frac{1}{2m^2 +N^2 -2mN} = \frac{1}{2m(m-N) +N^2}\ge \frac{1}{N^2}$$
Now $$\sum_{N=1}^\infty \sum_{m=1}^{N-1}\frac{1}{m^2 + (N-m)^2} \ge \sum_{N=1}^\infty \sum_{m=1}^{N-1}\frac{1}{N^2} = \sum_{N=1}^{\infty}\frac{N-1}{N^2}$$
Can you finish from here?
A: $$
\begin{align}
\sum_{m=1}^\infty\sum_{n=1}^m\frac1{m^2+n^2}
&\ge\sum_{m=1}^\infty\sum_{n=1}^m\frac1{2m^2}\\
&=\frac12\sum_{m=1}^\infty\frac1m\\
&=+\infty.
\end{align}
$$
A: $$\iint_{(1,+\infty)^2}\frac{dx\,dy}{x^2+y^2}\geq \int_{\varepsilon}^{\pi/2-\varepsilon}\int_{\rho_0}^{+\infty}\frac{1}{\rho}\,d\rho\,d\theta = +\infty $$
hence your series is divergent by comparison with a divergent integral.
As an alternative, let
$$ r_2(n) = \left|\{(a,b)\in\mathbb{Z}^2:a^2+b^2=n\}\right| $$
and $\mathbb{1}_S(n)$ the characteristic function of the non-zero integers which are sums of two squares.
By summation by parts and by Gauss circle problem
$$ \sum_{n=1}^{N}\frac{\mathbb{1}_S(n) r_2(n)}{n}=\pi+\sum_{n=1}^{N}\frac{\pi}{N}+O(1)=\pi\log(N)+O(1).$$
A: $\newcommand{\bbx}[1]{\,\bbox[15px,border:1px groove navy]{\displaystyle{#1}}\,}
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Note that

\begin{align}
\sum_{n = 1}^{\infty}{1 \over m^{2} + n^{2}} & =
\sum_{n = 1}^{\infty}{1 \over \pars{n - \ic m}\pars{n + \ic m}} =
{\Psi\pars{-\ic m} - \Psi\pars{\ic m} \over -2\ic m}
\\[5mm] & =
{1 \over m}\,\Im\Psi\pars{\ic m}\qquad\pars{~\Psi:\ Digamma\ Function~}
\\[5mm] & =
{1 \over 2m^{2}} + {\pars{\pi/2}\coth\pars{\pi m} \over m}
\,\,\,\stackrel{\mrm{as}\ m\ \to\ \infty}{\sim}\,\,\,{\pi \over 2}\,{1 \over m}
\end{align}

such that the double series diverges because
  $\ds{\sum_{m = 1}^{M}{\pi \over 2}\,{1 \over m} = {\pi \over 2}\,H_{M}}$ where $\ds{H_{z}}$ is the Harmonic Number.

See $\ds{\mathbf{\color{#000}{6.3.11}}}$ in A & S Table.
