Find the sum of the series $ \ \sum_{n=1}^{\infty} \frac{1}{n^2+l^2} \ $ Find the sum of the series $ \ \sum_{n=1}^{\infty} \frac{1}{n^2+l^2} \ $ , where $ \ l=constant \ $ 
Answer
The given series is convergent clearly . 
$ \ \sum_{n=1}^{\infty} \frac{1}{n^2+l^2} \\ = \frac{1}{1^2+l^2}+\frac{1}{2^2+l^2}+......+\frac{1}{n^2+l^2}+........ \\ = \frac{1}{l}  [ d \tan^{-1}(\frac{1}{l})+ d \tan^{-1}(\frac{2}{l})+........+ d \tan^{-1}(\frac{n}{l})+.....] \\ = \frac{1}{l} \lim_{n \to \infty}  \sum_{k=1}^{n}[d \tan^{-1}(\frac{1}{l})+ d \tan^{-1}(\frac{2}{l})+........+ d \tan^{-1}(\frac{k}{l})]    \\ = \frac{1}{l} \ d (\lim_{n \to \infty}  \sum_{k=1}^{n}[ \tan^{-1}(\frac{1}{l})+  \tan^{-1}(\frac{2}{l})+........+  \tan^{-1}(\frac{k}{l})]) $
but now i  can't proceed to find the sum of the series. 
Help me out.

Hint: Use the fact that
\begin{align}
\sum^\infty_{n=-\infty} \frac{1}{x^2+n^2} = \frac{\pi\coth(\pi x) }{x}.
\end{align}
