<s>[INST] 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.
 [/INST] 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, $S_o=\frac{3\pi\text{cot}(\frac{\pi}{3})-6\pi\text{cot}(\frac{2\pi}{3})}{72}=\frac{\sqrt{3}\pi}{24}$
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