<s>[INST] Proving $\sum_{n=1}^{\infty}\frac{1}{1+n^2\pi^2} = \frac{1}{e^2-1}$ I hope I'm allowed to ask this question here, but I have to prove that $\sum_{n=1}^{\infty}\frac{1}{1+n^2\pi^2} = \frac{1}{e^2-1}$ using the following Fourier series:
$$
  1-\frac{1}{e} + \sum_{n=1}^{\infty}\frac{2}{1+n^2\pi^2}\left[1-\frac{1}{e}(-1)^n\right]\cos(n{\pi}x) =
  \begin{cases} 
        e^x & x\in[-1,0) \\
        e^{-x} & x\in[0,1]
     \end{cases}
$$
This is my progress so far:
Let $x=0$:
\begin{align*}
  \therefore 1-\frac{1}{e} + \sum_{n=1}^{\infty}\frac{2}{1+n^2\pi^2}\left[1-\frac{1}{e}(-1)^n\right] &= 1 \\
  \sum_{n=1}^{\infty}\frac{2}{1+n^2\pi^2}\left[1-\frac{1}{e}(-1)^n\right] &= \frac{1}{e} \\
  \sum_{n=0}^{\infty}\frac{2}{1+(2n+1)^2\pi^2}\left(1+\frac{1}{e}\right) + \sum_{n=1}^{\infty}\frac{2}{1+(2n)^2\pi^2}\left(1-\frac{1}{e}\right) &= \frac{1}{e}
\end{align*}
Does anyone know what I should do next?
 [/INST] $\sum_{n=1}^{\infty}\frac{2}{1+n^2\pi^2}
=\frac{i}{\pi}\sum_{n=1}^{\infty}(\frac{1}{n+i/\pi}-\frac{1}{n-i/\pi})
=\frac{i(-\phi^{(0)}(\infty-i/\pi+1)+\phi^{(0)}(\infty+i/\pi+1)-\phi^{(0)}(1+i/\pi)+\phi^{(0)}(1-i/\pi))}{2\pi}=\frac{1}{2(e-1)}-\frac{1}{2(e+1)}=\frac{1}{e^2-1}$
where $\phi^{(0)}$ is the poly-gamma function of order $0$.
</s>