Evaluating $\sum\limits_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$ What is the value of $\displaystyle\sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$?
 A: Let's observe the relation with $\;\sin(2\pi j/3)$ and rewrite this in an elementary way :
\begin{align}
\sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}&= \sum_{k=1}^{\infty}\frac 0{3k+0}+\frac 1{3k+1}-\frac 1{3k+2}\\
&= \frac 2{\sqrt{3}}\Im\left(\sum_{j=1}^\infty\frac {e^{2\pi\,i\,j/3}}{j}\right)-\frac 1{1\cdot 2}\\
&= -\frac 2{\sqrt{3}}\Im\,\ln\left(1-e^{2\pi\,i/3}\right)-\frac 12\\
&= -\frac 2{\sqrt{3}}\Im\,\ln\left(\frac 32-\frac{\sqrt{3}}2i\right)-\frac 12\\
&= -\frac 2{\sqrt{3}}\Im\,\ln\left(\sqrt{3}\;e^{-\pi\,i/6}\right)-\frac 12\\
&= -\frac 2{\sqrt{3}}\left(-\frac{\pi}6\right)-\frac 12\\
&= \frac {\sqrt{3}\,\pi}{9}-\frac 12\\
\end{align}
A: \begin{align}
&\sum_{k = 0}^{\infty}{1 \over \left(3k + 1\right)\left(3k + 2\right)}
=
{1 \over 9}\sum_{k = 0}^{\infty}{1 \over \left(k + 1/3\right)\left(k + 2/3\right)}
=
{1 \over 9}\,{\Psi\left(1/3\right) - \Psi\left(2/3\right) \over 1/3 - 2/3}
\\[3mm]&=
{{1 \over 3}\left\lbrack \Psi\left(2 \over 3\right) - \Psi\left(1 \over 3\right)\right\rbrack}
=
{1 \over 3}\,\pi\ {\rm cotan}\left(\pi \over 3\right)
=
{1 \over 3}\,\pi\ {\sqrt{3} \over 3}
\\[5mm]&
\end{align}
$\Psi\left(z\right)$ is the Digamma function. In the last step I use
$\Psi\left(z\right) - \Psi\left(1 - z\right) = -\pi\,{\rm cotan}\left(\pi\,z\right)$ with
$z = 1/3$.
$$
\begin{array}{|c|}\hline\\
{\large\quad\sum_{k = 1}^{\infty}{1 \over \left(3k + 1\right)\left(3k + 2\right)}
=
\sum_{k = 0}^{\infty}{1 \over \left(3k + 1\right)\left(3k + 2\right)} - {1 \over 2}
=
{\sqrt{3} \over 9}\,\pi - {1 \over 2}\quad}
\\
\\
\hline
\end{array}
$$
A: Using $(7)$ from this answer,
$$
\begin{align}
\sum_{k=1}^\infty\frac1{(3k+1)(3k+2)}
&=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1{3k+1}-\frac1{3k+2}\right)\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1{3k+1}+\sum_{k=-n-1}^{-2}\frac1{3k+1}\right)\\
&=-\frac12+\lim_{n\to\infty}\sum_{k=-n}^n\frac1{3k+1}\\
&=-\frac12+\frac13\sum_{k=-\infty}^\infty\frac1{k+\frac13}\\
&=-\frac12+\frac\pi3\cot\left(\frac\pi3\right)\\
&=\frac\pi{3\sqrt3}-\frac12
\end{align}
$$
A: I like solution with Gamma/Beta function:
\begin{aligned}  \sum_{k=1}^{+\infty}\frac{1}{(3k+1)(3k+2)}  & =\sum_{k=1}^{+\infty} \frac{\Gamma(3k+1)}{\Gamma(3k+3)} \\& =\sum_{k=1}^{+\infty}\operatorname{B}(3k+1, 2) \\& = \sum_{k=1}^{+\infty}\int_{0}^{1}x^{3k}(1-x)\, dx\\&= \int_{0}^{1} \sum_{k=1}^{+\infty}x^{3k}(1-x)\, dx\\& = \int_{0}^{1} \frac{x^3}{1+x+x^2} \ dx  \\& = \frac{\sqrt{3}\pi}{9}-\frac{1}{2} \end{aligned}
