Simplification of arctan-like infinite series

Question

I'm trying to find a simplified expression for

$\sum_{n=0}^{\infty} (2n+1) (-1)^n \left[ \frac{a^2}{2 b^2} + \frac{c^2 \pi^2 (2n+1)^2}{2 d^2} \right]^{-1}$

For the special case $a=0$ I obtained

$\frac{2 d^2}{c^2 \pi^2} \sum_{n=0}^{\infty} \frac{(-1)^n}{2n+1} = \frac{2 d^2}{c^2 \pi^2} \text{arctan}(1) = \frac{2 d^2}{c^2 \pi^2} \frac{\pi}{4} = \frac{d^2}{2 c^2 \pi}$

Does anybody know if I can get a similar simplification for the general case? I have to evaluate this formula repeatedly so I'm looking for a solution that doesn't involve a series or that involves a series whose terms quickly approach zero.

Any help would be appreciated!

Answer 1

I think that a closed form is possible.

It can be proven, for instance using Fourier series for $t\mapsto\sin(\alpha t)$ that $$\frac{\pi}{4}\sec\left(\frac{\alpha\pi}{2}\right)=\sum_{n=0}^\infty\left(\frac{(-1)^n(2n+1)}{(2n+1)^2-\alpha^2}\right)$$ For instance, see this for a proof.

Replacing $\alpha$ with $i\alpha$: $$\frac{\pi}{4\cosh\left(\frac{\alpha \pi}{2}\right)}=\sum_{n=0}^\infty\frac{(-1)^n(2n+1)}{(2n+1)^2+\alpha^2}$$

The expression you're looking to simplify can be written in the following form:

$$\sum_{n=0}^{\infty} (2n+1) (-1)^n \left[ \frac{a^2}{2 b^2} + \frac{c^2 \pi^2 (2n+1)^2}{2 d^2} \right]^{-1} = \beta\sum_{n=0}^{\infty} \frac{(-1)^n(2n+1) }{ \alpha^2 + (2n+1)^2}$$ with $\beta=\frac{2d^2}{c^2\pi^2}$ and $\alpha=\frac{da}{\pi bc}$.

Therefore $$\sum_{n=0}^{\infty} (2n+1) (-1)^n \left[ \frac{a^2}{2 b^2} + \frac{c^2 \pi^2 (2n+1)^2}{2 d^2} \right]^{-1} = \frac{d^2}{2\pi c^2\cosh\left(\frac{da}{2bc}\right)}$$