An infinite series that converges to $\frac{\sqrt{3}\pi}{24}$ 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.
 A: Write the sum as $S_{o}=\sum_{n=1}^{\infty} \frac{1}{(3(2n+1)-2)(3(2n+1)+2)}=\sum_{n=\text{odd}} \frac{1}{(3n)^2-(2)^2}$.
Hence, $S_o=S-S_e$. Where, for $S$ , $n$ are all natural numbers in the above expression and for $S_e$, $n$ takes only positive even numbers.
Hence, from the identity $\frac{\pi\text{cot}(x)}{x}=\frac{1}{x^2}+\sum_{n=1}^{\infty} \frac{2}{x^2-n^2}$, we get $S=\frac{9-6\pi\text{cot}(\frac{2\pi}{3})}{72}$
And, $S_e=\frac{1}{4}\sum_{n=1}^{\infty} \frac{1}{(3n)^2-1}$. This similarly gives $S_e=\frac{9-3\pi\text{cot}(\frac{\pi}{3})}{72}$.
Hence, $S_o=\frac{3\pi\text{cot}(\frac{\pi}{3})-6\pi\text{cot}(\frac{2\pi}{3})}{72}=\frac{\sqrt{3}\pi}{24}$
A: Here is an elementary proof. We rewrite the series as
$$\frac{1}{4}\int_0^1\frac{1-x^4}{1-x^6}\,dx=\frac{1}{8}\int_0^1\frac{dx}{1-x+x^2}+\frac{1}{8}\int_0^1\frac{dx}{1+x+x^2}.$$
It is straightforward to show that
\begin{align*}
\int_0^1\frac{dx}{1-x+x^2}&=\frac{2\pi}{3\sqrt{3}},\\
\int_0^1\frac{dx}{1+x+x^2}&=\frac{\pi}{3\sqrt{3}},
\end{align*}
so we are done.
A: 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.
