An Alternating Sum

$S=\sum_{n \ \text{even}} \dfrac{1}{1+n^2}-\sum_{n \ \text{odd}} \dfrac{1}{1+n^2}$ Now, $\displaystyle\sum_{n \ \text{even}} \dfrac{1}{1+n^2}=\displaystyle\sum_{n=-\infty}^{\infty} \dfrac{1}{1+(2n)^2}=\dfrac{1}{4}\displaystyle\sum_{n=-\infty}^{\infty} \dfrac{1}{1/4+n^2}$ .

Also, $\begin{aligned}\sum_{n \ \text{odd}} \dfrac{1}{1+n^2}&=\sum_{n=-\infty}^{\infty}\dfrac{1}{1+n^2}-\sum_{n \ \text{even}} \dfrac{1}{1+n^2}\\ &=\sum_{n=-\infty}^{\infty}\dfrac{1}{1+n^2}-\dfrac{1}{4}\sum_{n=-\infty}^{\infty} \dfrac{1}{1/4+n^2}\end{aligned}$

Hence, $S=\dfrac{1}{2}\displaystyle\sum_{n=-\infty}^{\infty} \dfrac{1}{1/4+n^2}-\sum_{n=-\infty}^{\infty}\dfrac{1}{1+n^2} \ \ \ \ \ \ \ \ \ (1)$

Now, consider $f(x)=e^{-a|x|}, \ a>0$ . Let $F(k)$ be the Fourier transform of $f$ .

$\begin{aligned} \therefore \ F(k)&=\int_{-\infty}^{\infty}f(x)e^{-ikx}\mathrm{d}x\\ &= \int_{-\infty}^{\infty}e^{-a|x|}e^{-ikx}\mathrm{d}x\\ &=\dfrac{2a}{a^2+k^2}\end{aligned}$

Now according to the Poisson summation formula , $\sum_{n=-\infty}^{\infty}f(n)=\sum_{m=-\infty}^{\infty}F(m)$

Therefore, $\sum_{n=-\infty}^{\infty}e^{-a|n|}=\sum_{m=-\infty}^{\infty}\dfrac{2a}{a^2+(2\pi m)^2}$

The LHS is just a geometric series. Summing it, we get

$\begin{aligned}\sum_{m=-\infty}^{\infty}\dfrac{2a}{a^2+(2\pi m)^2}&=\dfrac{1+e^{-a}}{1-e^{-a}}\\ \sum_{m=-\infty}^{\infty} \dfrac{1}{(a/2\pi)^2+n^2}&=\dfrac{2\pi^2}{a}\dfrac{1+e^{-a}}{1-e^{-a}}\end{aligned}$

Put $a=2\pi,\pi/2$ and substitute the results into $(1)$ to obtain $\boxed{S=\dfrac{2\pi e^{\pi}}{e^{2\pi}-1}}$