TITLE: An infinite series that converges to $\frac{\sqrt{3}\pi}{24}$
QUESTION [8 upvotes]: Can you prove or disprove the following claim:

Claim:
$$\frac{\sqrt{3} \pi}{24}=\displaystyle\sum_{n=0}^{\infty}\frac{1}{(6n+1)(6n+5)}$$

The SageMath cell that demonstrates this claim can be found here.

REPLY [12 votes]: Let the sum be $S$. First of all, it is easy to see that
$$S=\frac{1}{2}\sum_{n=-\infty}^\infty f(n),\quad\mbox{where}\quad f(z)=\frac{1}{(6z+1)(6z+5)}.$$
This is true because $f(z)=f(-1-z)$.
Then by the summation formula (see any undergraduate Complex Variables textbook)
$$\sum_{n=-\infty}^\infty f(n)=-\sum_a{\mathrm{res}}_a \left(f(z)\pi\cot\pi z\right),$$
where summation is over all poles of $f$, that is $a_1=-1/6$ and $a_2=-5/6$.
Computing the residues (each of them equals $-\pi\sqrt{3}/24$) we obtain the result.