An infinite series that converges to $\frac{\sqrt{3}\pi}{24}$

Question

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.

Answer 1

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.