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So I want to compute $$\sum_{1}^\infty \frac{1}{n^{2}+1} $$ where $$y(x)=e^{x}, 0\le x \le 2\pi $$

I want to use Parsevals formula, and solving for the fourier coefficents I get that: $$c_{n}=\frac{1}{2\pi} * \int_{0}^{2\pi} e^{x(1-in)} dx = \frac{1}{2\pi}*\frac{e^{2\pi}-1}{1-in} $$ So $$ \sum_{-\infty}^\infty |c_{n}|^{2} = \frac{1}{4\pi^{2}}(e^{2\pi}-1)^{2}\sum_{-\infty}^\infty \frac{1}{1-2in-n^{2}} $$ And this is of course equal to the integral of the function squared...But my problem is that this sum on the right hand side is not on the form $n^{2}+1$, how can i make my numerator the same?

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You miscomputed $|c_n|^2$. Don't worry; this is common. Note that $|1 - in|^2 = 1+n^2$.

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    Im not sure if im following, isnt $ (1-in)^{2} = 1-2in-n^{2} $ ? – 2017-02-10
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    @fejz1234 what you are looking for is $|1-in|^2$, where $|\cdot|$ is the modulus of a complex number. – 2017-02-10
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    Oh I totally missed that absolute value... so $|1-in| = \sqrt{1^{2} + n^{2}}$ Hence $|1-in|^{2} = (\sqrt{1^{2} + n^{2}})^{2} = 1 + n^{2}$ ? – 2017-02-10
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    @fejz1234 yes. ${}$ – 2017-02-10