<s>[INST] Simplification of arctan-like infinite series 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!
 [/INST] 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
$$\sum_{n=0}^{\infty} (2n+1) (-1)^n \left[ \frac{a^2}{2 b^2} + \frac{c^2 \pi^2 (2n+1)^2}{2 d^2} \right]^{-1} = \frac{d^2}{2\pi c^2\cosh\left(\frac{da}{2bc}\right)}$$
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