Evaluating $\sum_{n=1}^{\infty} \frac{1}{n^2+1}$ While I know that $$\sum_{n=1}^{\infty} \frac{1}{n^2} = \frac{\pi^{2}}{6}$$
But trying to evaluate this has left me stumped
$$\sum_{n=1}^{\infty} \frac{1}{n^2+1}$$
I evaluated it through wolfram alpha, it gave me $\frac{1}{2}(\pi\coth(\pi)-1)$.
What would be a good way to start evaluating this series?

This sum is harder, and perhaps less natural, to find than the sum for $\sum \frac{1}{n^2}$. We use Fourier series of $e^{x}$ with respect to $\{e^{i nx} / 2\pi\}$and apply Parseval's theorem.
Note that we can use this method of evaluation to find $\sum \frac{1}{n^2}$, too! We do the same for the (simpler) function $f(x) = x$. We could also directly evaluate the Fourier series for $f(x) = (\pi - |x|)^2$ on $(-\pi, \pi)$ at $x = 0$ to conclude the same (this function aso gives us the sum $\sum \frac{1}{n^4} = \frac{\pi^4}{90}$). (Source of these examples to Rudin's Principles of Mathematical Analysis)
We begin by calculating $\displaystyle \int_{-\pi}^{\pi} |f(x)|^2 \, dx = \int_{-\pi}^{\pi} e^{2x} \, dx = \frac{e^{2\pi} - e^{-2\pi}}{2}$.
Moving on to the Fourier coefficients, we have $c_n = \displaystyle \dfrac{1}{\sqrt{2\pi}} \int_{-\pi}^{\pi} e^{-inx+x} \, dx = \dfrac{1}{\sqrt{2\pi}}\left( \int_{-\pi}^{\pi} e^x \cos(nx)\,dx - i\int_{-\pi}^{\pi} e^x \sin(nx)\,dx\right).$ Integrating the first integral with parts $f = \cos(nx)$ and $e^x\,dx = dg$ we get $$\left.e^x \cos(nx)\right|_{-\pi}^{\pi} + n\int_{-\pi}^{\pi} e^x \sin(nx)\,dx$$
We integrate by parts again and combine the resulting "like terms" (integrals) to get $$(n^2+1) \int_{-\pi}^{\pi} e^x \cos(nx) \, dx = e^x (\cos(nx) + n \sin(nx))|_{-\pi}^{\pi} = (-1)^n (e^{\pi} - e^{-\pi})$$
and so $$\frac{1}{\sqrt{2\pi}} \int_{-\pi}^{\pi} e^x \cos(nx) = \frac{(-1)^n}{\sqrt{2\pi}} \frac{e^{\pi} - e^{-\pi}}{n^2 + 1}.$$
Similarly, we get $$\dfrac{1}{\sqrt{2\pi}} \int_{-\pi}^{\pi} e^x \sin(nx) \, dx = -\frac{n(-1)^n}{\sqrt{2\pi}} \frac{e^{\pi} - e^{-\pi}}{n^2 + 1}.$$
Hence $$|c_n|^2 = \frac{(e^\pi - e^{-\pi})^2 + n^2 (e^{\pi} - e^{-\pi})^2}{2\pi(n^2+1)^2} = \frac{1}{2\pi}\frac{(e^{\pi}-e^{-\pi})^2}{n^2+1}$$ since $|a+bi|^2 = a^2 + b^2$.
Putting it all together using Parseval's formula we have
\begin{align} \sum_{n=-\infty}^{\infty} |c_n|^2 &=  \int_{-\pi}^{\pi} |f(x)|^2 \, dx \\
\frac{1}{2\pi} \sum_{n=-\infty}^{\infty} \frac{(e^\pi - e^{-\pi})^2}{n^2+1} &= \frac{e^{2\pi}-e^{-2\pi}}{2}\\
\sum_{n=-\infty}^{\infty} \frac{1}{n^2 + 1} &= \pi\frac{e^{2\pi} - e^{-2\pi}}{(e^\pi - e^{-\pi})^2}\\
2 \sum_{n=1}^{\infty} \frac{1}{n^2+ 1} &= \pi \frac{e^\pi + e^{-\pi}}{e^{\pi} - e^{-\pi}} - 1\\
\sum_{n=1}^{\infty} \frac{1}{n^2 + 1} &= \frac{\pi \coth \pi - 1}{2}
\end{align}
where we use $e^{2\pi} - e^{-2\pi} = (e^{\pi} - e^{-\pi})(e^{\pi}+e^{-\pi})$ and $c_n = c_{-n}$ in the third line. 
