TITLE: Evaluating $\sum\limits_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$
QUESTION [8 upvotes]: What is the value of $\displaystyle\sum_{k=1}^{\infty}\frac1{(3k+1)(3k+2)}$?

REPLY [3 votes]: Using $(7)$ from this answer,
$$
\begin{align}
\sum_{k=1}^\infty\frac1{(3k+1)(3k+2)}
&=\lim_{n\to\infty}\sum_{k=1}^n\left(\frac1{3k+1}-\frac1{3k+2}\right)\\
&=\lim_{n\to\infty}\left(\sum_{k=1}^n\frac1{3k+1}+\sum_{k=-n-1}^{-2}\frac1{3k+1}\right)\\
&=-\frac12+\lim_{n\to\infty}\sum_{k=-n}^n\frac1{3k+1}\\
&=-\frac12+\frac13\sum_{k=-\infty}^\infty\frac1{k+\frac13}\\
&=-\frac12+\frac\pi3\cot\left(\frac\pi3\right)\\
&=\frac\pi{3\sqrt3}-\frac12
\end{align}
$$