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.
Start with the infinite product expansion of $\cos x$
$$\cos x = \prod_{k=0}^\infty \left(1 - \frac{x^2}{(k+\frac12)^2\pi^2}\right)$$
Taking logarithm of $\cos(\pi x)$ and differentiate, we have
$$-\pi\tan(\pi x) = \sum_{k=0}^\infty \frac{2x}{x^2-(k+\frac12)^2} \quad\implies\quad \sum_{k=0}^\infty\frac{1}{(k+\frac12)^2-x^2} = \frac{\pi}{2x}\tan(\pi x) $$ This lead to $$ \sum_{k=0}^\infty \frac{1}{k^2+7k+9} = \sum_{k=0}^\infty \frac{1}{(k+\frac72)^2 - \frac{13}{4}} = \sum_{k=3}^\infty \frac{1}{(k+\frac12)^2 - \frac{13}{4}}\\ = \frac{\pi}{2\cdot\frac{\sqrt{13}}{2}}\tan\left(\frac{\pi\sqrt{13}}{2}\right) - \sum_{k=0}^2 \frac{1}{(k+\frac12)^2 - \frac{13}{4}} = 1 + \frac{\pi}{\sqrt{13}}\tan\left(\frac{\pi\sqrt{13}}{2}\right) $$
$$ \begin{align} \sum_{n=0}^\infty\frac1{n^2+7n+9} &=\frac1{\sqrt{13}}\sum_{n=0}^\infty\left(\frac1{n+\frac72-\frac{\sqrt{13}}2}-\frac1{n+\frac72+\frac{\sqrt{13}}2}\right)\\ &=\frac1{\sqrt{13}}\sum_{n=0}^\infty\left(\frac1{n+\frac72-\frac{\sqrt{13}}2}+\frac1{-n-\frac72-\frac{\sqrt{13}}2}\right)\\ &=\frac1{\sqrt{13}}\sum_{n\in\mathbb{Z}}\frac1{n+\frac{1}{2}-\frac{\sqrt{13}}{2}}\\ &-\frac1{\sqrt{13}}\left(\frac1{\frac52-\frac{\sqrt{13}}{2}}+\frac1{\frac32-\frac{\sqrt{13}}{2}}+\frac1{\frac12-\frac{\sqrt{13}}{2}}\right)\\ &-\frac1{\sqrt{13}}\left(\frac1{-\frac52-\frac{\sqrt{13}}{2}}+\frac1{-\frac32-\frac{\sqrt{13}}{2}}+\frac1{-\frac12-\frac{\sqrt{13}}{2}}\right)\\ &=\frac\pi{\sqrt{13}}\cot\left(\frac\pi2-\frac{\pi\sqrt{13}}{2}\right)-\frac1{\sqrt{13}}\left(\frac{\sqrt{13}}{3}-\frac{\sqrt{13}}{1}-\frac{\sqrt{13}}{3}\right)\\[3pt] &=\frac\pi{\sqrt{13}}\tan\left(\frac{\pi\sqrt{13}}{2}\right)+1 \end{align} $$ where we use $(7)$ from this answer.
Too long for a comment: The solution consists in using one of the following three formulas: $$\sum_{n=-\infty}^\infty\dfrac1{(n+a)(n+b)}=-\pi\cdot\frac{\cot(a\pi)-\cot(b\pi)}{a-b}$$ or $$\sum_{n=1}^\infty\dfrac1{(n+a)(n+b)}=\dfrac{H_a-H_b}{a-b}$$ or $$\sum_{n=1}^\infty\dfrac1{(n+a)^2-b^2}=\dfrac{H_{a+b}-H_{a-b}}{2b}$$ where $H_k$ is the $\big($generalized$\big)$ harmonic number, whose relationship to the digamma function has been well-studied, $~\psi(k+1)=H_k-\gamma$, where $\gamma\simeq\dfrac1{\sqrt3}$ is the famous Euler-Mascheroni constant.
Can't you use the first comparison test? Your series is always less than $\sum_{n=0}^\infty \frac{1}{n^2}$ which converges.
