Evaluating $\sum\limits_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$ What is the value of $\displaystyle\sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$?

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}
