Consider the following double sum
$$
S = \sum_{n=1}^\infty \sum_{m=1}^\infty
\frac{1}{a (2n-1)^2 – b (2m-1)^2} \, ,
$$
where $a$ and $b$ are both positive real numbers given by
\begin{align}
a &= \frac{1}{2} – \frac{\sqrt{2}}{32} \, , \\
b &= \frac{1}{4} – \frac{3\sqrt{2}}{32} \, .
\end{align}
It turns out that one of the two sums can readily be calculated and expressed in terms of the tangente function.
Specifically,
$$
S = \frac{\pi}{4\sqrt{ab}} \sum_{m=1}^\infty
\frac{\tan \left( \frac{\pi}{2} \sqrt{\frac{b}{a}} (2m-1) \right)}{2m-1} \, .
$$
The latter result does not seem to be further simplified. i was wondering whether someone here could be of help and let me know in case there exists a method to evaluate the sum above. Hints and suggestions welcome.
Thank you
PS From numerical evaluation using computer algebra systems, it seems that the series is convergent. This apparently would not be the case if $b<0$.
Best Answer
Answer: The series diverges when $\sqrt{\frac{b}{a}}$ is irrational, as in your case. This follows from your reduced form of the sum and a few observations:
Thus there are infinitely many terms in the sum which are bigger in magnitude than some positive constant, so the sum can't converge.
Remark: If $\sqrt{\frac{b}{a}}$ is some rational number $\frac{p}{q}$, then the convergence depends on the parity of $p$ and $q$. I'll sketch what happens in each case:
So summarily the sum converges only when $\sqrt{\frac{b}{a}}$ is a rational number with either the numerator or denominator even. When it does converge, you should be able to find an explicit formula for the sum by adding up a finite number of sums of the form $\sum \frac{1}{a+n^2}$ (arising from the pairing of points mentioned above), which can each be evaluated analytically.
(This is quite sketchy, but it doesn't apply to your particular value anyway, so I'm hoping I can get away with it...)