Find $S=\sum_{n=-\infty}^{\infty} \frac{(-1)^n}{1+n^2}$ How do I find the sum:
$$S=\sum_{n=-\infty}^{\infty} \dfrac{(-1)^n}{1+n^2}$$
I can't solve this can someone help me?
 A: $$S=\sum_{n \ \text{even}} \dfrac{1}{1+n^2}-\sum_{n \ \text{odd}} \dfrac{1}{1+n^2}$$
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,
$$\begin{align}\sum_{n \ \text{odd}} \dfrac{1}{1+n^2}&=\sum_{n=-\infty}^{\infty}\dfrac{1}{1+n^2}-\sum_{n \ \text{even}} \dfrac{1}{1+n^2}\\ &=\sum_{n=-\infty}^{\infty}\dfrac{1}{1+n^2}-\dfrac{1}{4}\sum_{n=-\infty}^{\infty} \dfrac{1}{1/4+n^2}\end{align}$$
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 $f$.
$$\begin{align} \therefore \ F(k)&=\int_{-\infty}^{\infty}f(x)e^{-ikx}\mathrm{d}x\\ &= \int_{-\infty}^{\infty}e^{-a|x|}e^{-ikx}\mathrm{d}x\\ &=\dfrac{2a}{a^2+k^2}\end{align}$$
Now according to the Poisson summation formula,
$$\sum_{n=-\infty}^{\infty}f(n)=\sum_{m=-\infty}^{\infty}F(2\pi m)$$
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
$$\begin{align}\sum_{m=-\infty}^{\infty}\dfrac{2a}{a^2+(2\pi m)^2}&=\dfrac{1+e^{-a}}{1-e^{-a}}\\ \sum_{m=-\infty}^{\infty} \dfrac{1}{(a/2\pi)^2+n^2}&=\dfrac{2\pi^2}{a}\dfrac{1+e^{-a}}{1-e^{-a}}\end{align}$$
Put $a=2\pi,\pi/2$ and substitute the results into $(1)$ to obtain
$$\therefore{S=\dfrac{2\pi e^{\pi}}{e^{2\pi}-1}}$$
A: We can use a result from complex analysis, specifically the residue theorem:
$$\sum_{n=-\infty}^{\infty} (-1)^n f(n) = -\pi \sum_k \operatorname*{Res}_{z=z_k} [f(z) \csc{\pi z}] $$
where the $z_k$ are the poles of $f$.  here $f(z) = 1/(z^2+1)$.  The poles are at $z_{\pm} = \pm i$.  Thus the sum is
$$-\pi \frac1{2 i} \frac1{\sin{i \pi}} - \pi \frac1{-2 i} \frac1{\sin{(-i \pi)}} = \frac{\pi}{\sinh{\pi}} $$
A: Hint: Differentiate the natural logarithm of Euler's infinite product expression for the sine function, and then employ the well-known relations between trigonometric and hyperbolic functions, in order to show that $~\displaystyle\sum_{n=-\infty}^\infty\frac1{a^2+n^2} ~=~ \frac\pi a~\coth(\pi a),~$ then, using Yagna Patel's relation $(1)$, evaluate S.
