Finding $\sum \frac{1}{n^2+7n+9}$ 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
$$
\sum_{k=0}^\infty \frac{1}{k^2+7k+9}
= \sum_{k=0}^\infty \frac{1}{(k+\frac72)^2 - \frac{13}{4}}
= \sum_{k=3}^\infty \frac{1}{(k+\frac12)^2 - \frac{13}{4}}\\
= \frac{\pi}{2\cdot\frac{\sqrt{13}}{2}}\tan\left(\frac{\pi\sqrt{13}}{2}\right) - \sum_{k=0}^2 \frac{1}{(k+\frac12)^2 - \frac{13}{4}}
= 1 + \frac{\pi}{\sqrt{13}}\tan\left(\frac{\pi\sqrt{13}}{2}\right)
$$
Can't you use the first comparison test? Your series is always less than $\sum_{n=0}^\infty \frac{1}{n^2}$ which converges.