<s>[INST] The Sum of the series $\sum\limits_{n=0}^{\infty}\frac{1}{n^2+3}$ I know how to get the sum of geometric series, but otherwise.
How do I get the sum of this series?  Thank you.
$$\sum\limits_{n=0}^{\infty}\frac{1}{n^2+3}$$
 [/INST] 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.
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