TITLE: Finding $\sum \frac{1}{n^2+7n+9}$
QUESTION [13 upvotes]: How do we prove that $$\sum_{n=0}^{\infty} \dfrac{1}{n^2+7n+9}=1+\dfrac{\pi}{\sqrt {13}}\tan\left(\dfrac{\sqrt{13}\pi}{2}\right)$$
I tried partial fraction decomposition, but it didn't work out after that. Please help me out. Hints and answers appreciated. Thank you.

REPLY [6 votes]: $$
\begin{align}
\sum_{n=0}^\infty\frac1{n^2+7n+9}
&=\frac1{\sqrt{13}}\sum_{n=0}^\infty\left(\frac1{n+\frac72-\frac{\sqrt{13}}2}-\frac1{n+\frac72+\frac{\sqrt{13}}2}\right)\\
&=\frac1{\sqrt{13}}\sum_{n=0}^\infty\left(\frac1{n+\frac72-\frac{\sqrt{13}}2}+\frac1{-n-\frac72-\frac{\sqrt{13}}2}\right)\\
&=\frac1{\sqrt{13}}\sum_{n\in\mathbb{Z}}\frac1{n+\frac{1}{2}-\frac{\sqrt{13}}{2}}\\
&-\frac1{\sqrt{13}}\left(\frac1{\frac52-\frac{\sqrt{13}}{2}}+\frac1{\frac32-\frac{\sqrt{13}}{2}}+\frac1{\frac12-\frac{\sqrt{13}}{2}}\right)\\
&-\frac1{\sqrt{13}}\left(\frac1{-\frac52-\frac{\sqrt{13}}{2}}+\frac1{-\frac32-\frac{\sqrt{13}}{2}}+\frac1{-\frac12-\frac{\sqrt{13}}{2}}\right)\\
&=\frac\pi{\sqrt{13}}\cot\left(\frac\pi2-\frac{\pi\sqrt{13}}{2}\right)-\frac1{\sqrt{13}}\left(\frac{\sqrt{13}}{3}-\frac{\sqrt{13}}{1}-\frac{\sqrt{13}}{3}\right)\\[3pt]
&=\frac\pi{\sqrt{13}}\tan\left(\frac{\pi\sqrt{13}}{2}\right)+1
\end{align}
$$
where we use $(7)$ from this answer.