Question: What is the value of $$\sum\_{n=1}^{\infty}\frac1{1+n^2}$$
I don't know how to solve it.  

I tried to solve it by power series, but I failed.

Answer: 
Hint:
$$\frac{\pi x}{2}\coth \pi x=\frac{1}{2}+\sum\_{n=1}^{\infty }\frac{1}{\left ( \frac{n}{x} \right )^2+1}$$
at $x=1$
$$\sum\_{n=1}^{\infty }\frac{1}{n^2+1}=\frac{\pi}{2}\coth \pi -\frac{1}{2}$$
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\begin{align}
\sum\_{n = 1}^{\infty}{1 \over n^{2} + 1} & =
{1 \over 2\ic}\sum\_{n = 0}^{\infty}\pars{{1 \over n + 1 - \ic} -
{1 \over n + 1 + \ic}}
\\[5mm] & =
{1 \over 2\ic}\bracks{\vphantom{\Large A}\Psi\pars{1 + \ic} - \Psi\pars{1 - \ic}}\qquad
\pars{~\Psi:\ Digamma\ Function~}
\\[5mm] & =
{1 \over 2\ic}\braces{\vphantom{\LARGE A}%
{1 \over \ic} - \bracks{\vphantom{\Large A}\Psi\pars{1 - \ic} - \Psi\pars{\ic}}}\qquad
\pars{~\Psi\ Recurrence~}
\\[5mm] & =
-\,{1 \over 2} - {1 \over 2\ic}\,
\bracks{\vphantom{\large A}\pi\cot\pars{\pi\ic}}\qquad
\pars{~Euler\ Reflection\ Formula~}
\\[5mm] & =
-\,{1 \over 2} - {\pi \over 2\ic}\,\bracks{\vphantom{\large A}-\ic\coth\pars{\pi}} =
\bbx{\ds{\pi\coth\pars{\pi} - 1 \over 2}}
\end{align}