The Sum of the series $\sum\limits_{n=0}^{\infty}\frac{1}{n^2+3}$

Question

I know how to get the sum of geometric series, but otherwise.

How do I get the sum of this series? Thank you.

$$\sum\limits_{n=0}^{\infty}\frac{1}{n^2+3}$$

Answer 1

The Weierstrass product for the sine function gives: $$ \forall z\in\mathbb{C},\quad\sin(z) = z\prod_{n\geq 1}\left(1-\frac{z^2}{n^2 \pi^2}\right)\tag{1} $$ hence by replacing $z$ with $iz$ we get: $$ \forall z\in\mathbb{C},\quad\sinh(z) = \frac{e^z-e^{-z}}{2} = z\prod_{n\geq 1}\left(1+\frac{z^2}{n^2 \pi^2}\right)\tag{2} $$ and by considering the logarithmic derivative, given by $\frac{f'(z)}{f(z)}=\frac{d}{dz}\,\log f(z)$, we have: $$ \forall z\in\mathbb{C},\quad\coth(z)=\frac{e^z+e^{-z}}{e^{z}-e^{-z}}=\frac{1}{z}+\sum_{n\geq 1}\frac{2z}{z^2+n^2\pi^2}\tag{3} $$ and by replacing $z$ with $\pi\sqrt{3}$ it follows that: $$ \coth(\pi\sqrt{3})=\frac{e^{2\pi\sqrt{3}}+1}{e^{2\pi\sqrt{3}}-1}=\frac{1}{\pi\sqrt{3}}+\sum_{n\geq 1}\frac{2\pi\sqrt{3}}{\pi^2(n^2+3)}\tag{4} $$ so, by rearranging:

$$ \color{red}{\sum_{n\geq 0}\frac{1}{n^2+3}}=\frac{1}{6}\left(1+\pi\sqrt{3}\coth(\pi\sqrt{3})\right)=\color{red}{\frac{1}{6}+\frac{\pi\sqrt{3}}{6}\cdot\frac{e^{2\pi\sqrt{3}}+1}{e^{2\pi\sqrt{3}}-1}}\approx 1.0736.\tag{5} $$

Another approach (a Fourier-analytic one) is shown in this similar question.