Summation of series involving $\sinh$ of a square root Consider the following series:
$$
S = \sum_{\text{odd } n} \sum_{\text{odd } m} \frac{(-1)^{(n+m)/2}}{nm} \frac{\sinh( \pi \sqrt{n^2 + m^2}/2)}{\sinh( \pi \sqrt{n^2 + m^2})}
$$
From the physical context, one can argue that the series should converge to
$$
S = - \frac{\pi^2}{96},
$$
and numerically calculating the first 10–20 terms of this series seems to show rapid convergence to this value.  
Is it possible to prove this statement without appealing to a physical argument?  I honestly have no idea how to begin tackling a problem like this.
 A: Here is a solution that I have found while working on other lattice sums.
It utilizes a very simple result:

Define $f$ by 
  $$f(x)=\sum_{n=0}^{\infty} (-1)^n (2n+1) e^{-\pi x (n+\frac12)^2}.$$
  Then 
  $$\int_0^{\infty} e^{-y x} f(x)dx=\operatorname{sech}\sqrt{\pi y}.\tag{$\star$}$$

The proof is simple - just integrate term by term, and the result follows from the series 
$$\sum_{n=0}^{\infty} \frac{(-1)^n (2n+1)}{(2n+1)^2+(2x)^2}=\frac{\pi}{4} \operatorname{sech} \pi x.$$
Therefore, 
$$\large \operatorname{sech} \frac{\pi}{2} \sqrt{n^2+m^2} = \int_0^{\infty} e^{-\frac{\pi}{4} x(n^2+m^2)} f(x)dx.$$
Now put $(2n+1)$ and $(2m+1)$ instead of $n$ and $m$, multiply by $\displaystyle \, \frac{(-1)^{n+m}}{(2n+1)(2m+1)},$ and
sum both $n$ and $m$ over the non-negative integers:
$$ 2 S=\sum_{n,m=0}^{\infty} \frac{(-1)^{n+m}}{(2n+1)(2m+1)} \operatorname{sech}\left(\frac{\pi}{2} \sqrt{(2n+1)^2+(2m+1)^2}\right)
= \int_0^{\infty} f(x) \left(\sum_{n=0}^{\infty} \frac{(-1)^n e^{-\pi x (n+\frac12)^2}}{2n+1} \right)^2 dx$$
Here comes the neat part: we notice that
$$\frac{d}{dx}  \sum_{n=0}^{\infty} \frac{(-1)^n e^{-\pi x (n+\frac12)^2}}{2n+1} = -\frac{\pi}{4} f(x),$$
and conclude that 
$$ 2 S = \frac13 \left(-\frac{4}{\pi}\right)  \left(\sum_{n=0}^{\infty} \frac{(-1)^n e^{-\pi x (n+\frac12)^2}}{2n+1} \right)^3\Biggr{|}_0^\infty
\\= \frac13 \left(\frac{4}{\pi}\right) \left(\frac{\pi}{4}\right)^3=\frac{\pi^2}{48}.$$
Can you see how nicely it generalizes to higher dimension versions of this sum? 
Note: 
The function $f(x)$ is actually $\eta^3(i x)$ where $\eta$ is the Dedekind eta function (This is a consequence of Jacobi's triple product identity). The result $(\star)$
was obtained by Glasser, along with other beautiful formulas for integrals involving $\eta(i x)$, in his article 
Some Integrals of the Dedekind Eta-function (2008) (arxiv link). I omitted this detail simply becuase it is not important for 
this purpose.
