<s>[INST] Evaluating $\sum\limits_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$ What is the value of $\displaystyle\sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$?
 [/INST] \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
$z = 1/3$.
$$
\begin{array}{|c|}\hline\\
{\large\quad\sum_{k = 1}^{\infty}{1 \over \left(3k + 1\right)\left(3k + 2\right)}
=
\sum_{k = 0}^{\infty}{1 \over \left(3k + 1\right)\left(3k + 2\right)} - {1 \over 2}
=
{\sqrt{3} \over 9}\,\pi - {1 \over 2}\quad}
\\
\\
\hline
\end{array}
$$
</s>