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.

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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.

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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.

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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.

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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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An Introduction to the Operations with Seriesby Isaac Joachim Schwatt (1924; reprinted with corrections by Chelsea Publishing Company in 1962). A google-books .pdf file is freely available where I'm at. $\endgroup$1more comment