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.
 A: Actually the notation $\sum_{(m,n)\in\mathbb{Z}_+^2}$ does not make sense here, since the given series is not absolutely convergent. On the other hand,
$$ \sum_{m\in\mathbb{Z}^+}\sum_{n\in\mathbb{Z}^+}\frac{m^2-n^2}{(m^2+n^2)^2}=\sum_{m\geq 1}\left(-\frac{1}{2m^2}+\frac{\pi^2}{2\sinh^2(m\pi)}\right) $$
$$ \sum_{n\in\mathbb{Z}^+}\sum_{m\in\mathbb{Z}^+}\frac{m^2-n^2}{(m^2+n^2)^2}=-\sum_{n\geq 1}\left(-\frac{1}{2n^2}+\frac{\pi^2}{2\sinh^2(n\pi)}\right) $$
by applying $\frac{d^2}{dx^2}\log(\cdot)$ to the Weierstrass product for the sine function.
The identity $\sum_{n\geq 1}\frac{1}{\sinh^2(\pi n)}=\frac{1}{6}-\frac{1}{2\pi}$ is pretty well-known and it can be derived through the residue theorem and/or modular forms (see Zucker, The Summation of Series of Hyperbolic Functions) and/or the Poisson summation formula. In particular
$$ \sum_{m\in\mathbb{Z}^+}\sum_{n\in\mathbb{Z}^+}\frac{m^2-n^2}{(m^2+n^2)^2}=\color{red}{-\frac{\pi}{4}}.$$
A: 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.
A: 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.
