Evaluate $ \sum_{m > n > 0} \frac{m^2 - n^2}{(m^2 + n^2)^2} $ I have been trying around different ranges of summations:
$$ \sum_{(m,n) \in \mathbb{Z}^2} \frac{m^2 - n^2}{(m^2 + n^2)^2} = 0$$
That's not any good.  What about if we restrict to $m, n \in \mathbb{Z}$ as positive integers.  
$$ \sum_{m > 0, n > 0} \frac{m^2 - n^2}{(m^2 + n^2)^2} = 0$$
Now here's an anti-symmetry.  I do not like taking out the symmetry, but perhaps I can ask about: 
$$ \sum_{m >  n > 0} \frac{m^2 - n^2}{(m^2 + n^2)^2}  = \; ? \tag{$*$} $$
There doesn't seem to be a change of variables that can work.  And we've used symmetry about as much as we can.  This looks related to:
$$  \sum \frac{1}{n^2} = \frac{\pi^2}{6}$$
Perhaps this other series ($*$) also has a special value.

Fix $m>0$ for a moment and consider the sum
$$
H(m):=\sum_{n=0}^{\lfloor m/2\rfloor}\frac{m^2-n^2}{(m^2+n^2)^2}.
$$
Here the numerator of the term, call it $x(n,m)$, is at least $3m^2/4$, and the denomimator is at most $4m^4$. Therefore
$$
x(n,m)\ge\frac{3m^2/4}{4m^4}=\frac{3}{16m^2}
$$
in this range. There are at least $m/2$ terms in $H(m)$, so we arrive at the lower bound
$$
H(m)\ge \frac{3}{32m}.
$$
This shows that the double sum
$$
\sum_{m>n>0}\frac{m^2-n^2}{(m^2+n^2)^2}
$$
diverges.
