Emerson Barnes

2022-01-24

What is the value of
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\left(3k+1\right)\left(3k+2\right)}$

Armani Dyer

I like solution with Gamma/Beta function:
$\sum _{k=1}^{+\mathrm{\infty }}\frac{1}{\left(3k+1\right)\left(3k+2\right)}=\sum _{k=1}^{+\mathrm{\infty }}\frac{\mathrm{\Gamma }\left(3k+1\right)}{\mathrm{\Gamma }\left(3k+3\right)}$
$=\sum _{k=1}^{+\mathrm{\infty }}B\left(3k+1,2\right)$
$=\sum _{k=1}^{+\mathrm{\infty }}{\int }_{0}^{1}{x}^{3k}\left(1-x\right)dx$
$={\int }_{0}^{1}\sum _{k=1}^{+\mathrm{\infty }}{x}^{3k}\left(1-x\right)dx$
$={\int }_{0}^{1}\frac{{x}^{3}}{1+x+{x}^{2}}dx$
$=\frac{\sqrt{3}\pi }{9}-\frac{1}{2}$

spelkw

Let's observe the relation with $\mathrm{sin}\left(2\pi \frac{j}{3}\right)$ and rewrite this in an elementary way :
$\sum _{k=1}^{\mathrm{\infty }}\frac{1}{\left(3k+1\right)\left(3k+2\right)}=\sum _{k=1}^{\mathrm{\infty }}\frac{0}{3k+0}+\frac{1}{3k+1}-\frac{1}{3k+2}$
$=\frac{2}{\sqrt{3}}I\left(\sum _{j=1}^{\mathrm{\infty }}\frac{{e}^{2\pi i\frac{j}{3}}}{j}\right)-\frac{1}{1\cdot 2}$
$=-\frac{2}{\sqrt{3}}I\mathrm{ln}\left(1-{e}^{2\pi \frac{i}{3}}\right)-\frac{1}{2}$
$=-\frac{2}{\sqrt{3}}I\mathrm{ln}\left(\sqrt{3}{e}^{-\pi \frac{i}{6}}\right)-\frac{1}{2}$
$=-\frac{2}{\sqrt{3}}\left(-\frac{\pi }{6}\right)-\frac{1}{2}$
$=\frac{\sqrt{3}\pi }{9}-\frac{1}{2}$

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