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.

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.
