How to calculate the value of $\sum\limits_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}$? How do I calculate the value of the series $$\sum_{k=0}^{\infty}\frac{1}{(3k+1)\cdot(3k+2)\cdot(3k+3)}= \frac{1}{1\cdot2\cdot3}+\frac{1}{4\cdot5\cdot6}+\frac{1}{7\cdot8\cdot9}+\cdots?$$

By making use of the integral
$$\int_{0}^{1} \frac{(1-x)^2}{1-x^3} \, dx = \frac{1}{2} \, \left(\frac{\pi}{\sqrt{3}} - \ln 3 \right)$$
one can take the following path.
\begin{align}
S &= \sum_{k=0}^{\infty} \frac{1}{(3k+1)(3k+2)(3k+3)} \\
&= \sum_{k=0}^{\infty} \frac{\Gamma(3k+1)}{\Gamma(3k+4)} = \frac{1}{2} \, \sum_{k=0}^{\infty} B(3, 3k+1),
\end{align}
where $B(n,m)$ is the Beta function, which leads to
\begin{align}
S &= \frac{1}{2} \, \sum_{k=0}^{\infty} \, \int_{0}^{1} t^{2} \, (1-t)^{3k} \, dt \\
&= \frac{1}{2} \, \int_{0}^{1} \frac{t^{2} \, dt}{1- (1-t)^{3}} \\
&= \frac{1}{2} \, \int_{0}^{1} \frac{(1-x)^{2} \, dx}{1- x^3} \hspace{15mm} x = 1 - t \\
&= \frac{1}{4} \, \left(\frac{\pi}{\sqrt{3}} - \ln 3 \right). 
\end{align}
