Show that $\sum_{n=0}^\infty\frac{(-1)^n}{(2 n+1) \left(\frac{x^2}{(2 n+1)^2}+1\right)}=\frac{\pi}{4\cosh\left(\frac{\pi x}{2}\right)}$

Question

Is it true that

$$\sum_{n=0}^\infty\frac{(-1)^n}{(2 n+1) \left(\frac{x^2}{(2 n+1)^2}+1\right)}=\frac{\pi}{4\cosh\left(\frac{\pi x}{2}\right)}$$?

How can I prove it ?

Thanks for helping.

Answer 1

$\frac{1}{\cos z}$ has simple poles at $\pi\mathbb{Z}+\frac{\pi}{2}$. In particular, for any $n\in\mathbb{Z}$:

$$ \text{Res}\left(\frac{1}{\cos z},z=\frac{(2n-1)\pi}{2}\right)=(-1)^n \tag{1}$$ hence: $$ \frac{1}{\cos z}=\sum_{n\in\mathbb{Z}}\frac{(-1)^n}{z-\frac{(2n-1)\pi}{2}}\tag{2}$$ and by putting together the contributes given by $n$ and $1-n$: $$ \frac{1}{\cos z} = \sum_{m\geq 0}\frac{(-1)^{m+1}(2m+1)\pi}{z^2-\frac{(2m+1)^2\pi^2}{4}}.\tag{3} $$ Your claim follows by replacing $z$ with $\frac{\pi i x}{2}$ in $(3)$.

Answer 2

The sum is ready-made for the residue theorem. Recall that

$$\sum_{n=-\infty}^{\infty} (-1)^n f(n) = -\pi \sum_k \operatorname*{Res}_{z=z_k} \left [ \csc{(\pi z)} \, f(z) \right ] $$

where $z_k$ are the non-integer poles of $f$. Here,

$$f(z)= \frac{2 z+1}{x^2+(2 z+1)^2} = \frac12 \frac{z+1/2}{(z+1/2)^2+x^2/4}$$

Thus, the poles are at $z_{\pm} = -1/2 \pm i x/2$ and are simple. Thus, the sum is equal to

$$-\frac{\pi}{2} \cdot\frac12 \left (\frac1{2 \sin{(-\pi/2 + i \pi x/2)}} + \frac1{2 \sin{(-\pi/2 - i \pi x/2)}} \right ) = \frac{\pi}{4} \operatorname{sech}{\left ( \frac{\pi x}{2}\right )}$$