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.

Consider:
\begin{align}
\int_{0}^{\infty} e^{-s t} \, t \, dt &= \frac{1}{s^{2}} \\
S_{1} &= \sum_{n=1}^{\infty} e^{- n^{2} t} = \frac{\theta_{3}(0, e^{- t}) - 1}{2} \\
S_{2} &= \sum_{n=1}^{\infty} n^{2} \, e^{- n^{2} t} = \frac{\theta^{'}_{3}(0, e^{- t})}{2}
\end{align}
then
\begin{align}
\sum_{n,m=1}^{\infty} \frac{m^{2} - n^{2}}{(m^{2} + n^{2})^{2}} &= \int_{0}^{\infty} t \, \left[ S_{1} \cdot S_{2} - S_{1} \cdot S_{2} \right] \, dt = 0.
\end{align}
Verification, numerically, was obtained by use of Mathematica by calculating the first 10000 terms of the series and obtained $0$. 
For $m \geq 1$, $1 \leq n \leq m$, the series is divergent, and is also the case for $m > n$ in other forms.
