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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$\begingroup$ I think it should be possible to rewrite the sum in terms of a doubly infinite sum with n ranging over $\mathbb{Z}$. From there, I think the methods of math.stackexchange.com/questions/1322086/… will get you to a closed form. $\endgroup$– Kevin VentulloCommented Aug 2, 2021 at 3:54
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15$\begingroup$ Write this as $\frac{1}{4} \sum_{n = 0}^{\infty} \left(\frac{1}{6n + 1} - \frac{1}{6n + 5}\right) = \frac{1}{4} \sum_{n = 1}^{\infty} \frac{\chi_6(n)}{n} = \frac{1}{4} L(1,\chi_6)$, where $\chi_6$ is the quadratic Dirichlet character modulo $6$, which satisfies $\chi_6(n) = 1$ if $n \equiv 1 \pmod{6}$, $\chi_6(n) = -1$ if $n \equiv 5 \pmod{6}$, and $\chi_6(n) = 0$ if $(n,6) \neq 1$. Then Dirichlet's class number formula implies the result. $\endgroup$– Peter HumphriesCommented Aug 2, 2021 at 3:58
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1$\begingroup$ Another idea, related to @PeterHumphries answer, is to write his first step as the limit $s\to1$ of two Hurwitz zeta functions, and then use the known expression of this limit in terms of the digamma function evaluated at $1/6$ and $5/6$, which are related to $\sqrt{3}\pi$. $\endgroup$– efsCommented Aug 2, 2021 at 4:09
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1$\begingroup$ FYI, a large number of such summations, nearly all of which are much more complicated, can be found in Chapters IX-X (pp. 147-200) of An Introduction to the Operations with Series by 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. $\endgroup$– Dave L RenfroCommented Aug 2, 2021 at 13:56
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2$\begingroup$ Yet one more way, close to GH from MO's answer: Let $\omega$ be a primitive $6$-th root of unity and write this as $\sum_{n=1}^{\infty} \tfrac{\omega^n + \omega^{2n} - \omega^{4n} - \omega^{5n}}{n (8 \sqrt{-3})}$. Then write this as the limit as $r \to 1^-$ of $\tfrac{-1}{8 \sqrt{-3}} \left( \log(1-r \omega) + \log(1-r \omega^2) - \log(1- r \omega^4) - \log(1-r \omega^5) \right)$. (There are probably some typos in this.) $\endgroup$– David E SpeyerCommented Aug 2, 2021 at 17:49
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3 Answers
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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.
answered Aug 2, 2021 at 4:49
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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.
answered Aug 2, 2021 at 5:13
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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, $S_o=\frac{3\pi\text{cot}(\frac{\pi}{3})-6\pi\text{cot}(\frac{2\pi}{3})}{72}=\frac{\sqrt{3}\pi}{24}$
