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?

Hint:
Try to see what happens for $x=1$.
Explanation:
Your progress so far has indicated that splitting the sum into even and odd sums may be useful. To that extend you could define
$$A=\sum_{n=0}^\infty \frac{2}{1+(2n+1)^2 \pi^2}$$ and
$$B=\sum_{n=1}^\infty \frac{2}{1+(2n)^2 \pi^2}.$$
Now, you already have one linear equation for $A,B$, which is (from your progress so far):
$$\left(1+\frac1e\right) \cdot A + \left(1-\frac1e\right)\cdot B = \frac1e$$
or, multiplying the whole equation by $e$ to avoid fractions, $$(e+1)A + (e-1)B = 1$$
Now, all you need is a second linear equation involving $A$ and $B$. You can get that by taking $x=1$, since the cosine factor expands nicely into $\cos(n\pi x) = \cos (n\pi) = (-1)^n$
Guidelines:
You should really try to solve the problem on your own from here on. But if you want to make sure you are still on the right track:

*

*the second equation you get should be something along the lines of $$(e+1)A + (e-1)B = 1$$

*The solution to the system of equations should then be $A=\frac{e-1}{2(e+1)}, B=\frac{3-e}{2(e-1)}$

*The final sum then comes out to $A+B=\frac{2}{e^2-1}$.

