<s>[INST] The calculation of a series Calculate the series
\begin{equation}
\sum_{n=0}^{\infty} \dfrac{1}{(4n+1)(4n+3)}.
\end{equation}
 [/INST] We let 
$$f(z)=\dfrac{1}{(4z+1)(4z+3)}$$
There are two poles of $f(z)$.  They are $4z_0+3=0 \implies z_0=-\frac{3}{4}$ and $4z_1+1=0 \implies z_1=-\frac{1}{4}$.
Residue calculus tells us that
$$\sum_{n=-\infty}^{\infty} \dfrac{1}{(4n+1)(4n+3)}=-(\text{sum of residues of }\pi\cot(\pi z)f(z))$$
We calculate the residues of the poles ($b_0$ and $b_1$ for $z_0$ and $z_1$ respectively):
$$b_0=\operatorname {Res}_{z=z_0}=\lim_{z \to z_0}\frac{(z-z_0)\pi\cot (\pi z)}{(4z+1)(4z+3)}$$
Using L'Hopital's rule we have
$$b_0=\lim_{z \to z_0} \frac{\pi\cot (\pi z)-(z-z_0)\pi^2 \csc^2 (\pi z)}{4((4z+1)+(4z+3))}=\frac {\pi \cot (-3\pi/4)}{4(4)}=-\frac{\pi}{8}$$
Similarly, we have 
$$b_1=\lim_{z \to z_1} \frac{\pi\cot (\pi z)-(z-z_1)\pi^2 \csc^2 (\pi z)}{4((4z+1)+(4z+3))}=\frac {\pi \cot (-\pi/4)}{4(4)}=-\frac{\pi}{8}$$
So
$$
\sum_{n=-\infty}^{\infty} \dfrac{1}{(4n+1)(4n+3)}=-(-\frac{\pi}{8}-\frac{\pi}{8})=\frac{\pi}{4}\implies 
\sum_{n=0}^{\infty} \dfrac{1}{(4n+1)(4n+3)}=\frac{1}{2}\frac{\pi}{4}=\frac{\pi}{8}
$$
QED
</s>