Find the sum of series $\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}$ How to find the sum of series $\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}$ in the formal way? Numerically its value is $\approx 0.0217326$ and the partial sum formula contains the first derivative of the gamma function (by WolframAlpha).

There is an alternate method and is as follows.
Notice that 
$$ \frac{1}{n(n+1)(n+2)} = \frac{(n-1)!}{(n+2)!} = \frac{1}{2!} \, B(n,3) $$
where $B(x,y)$ is the Beta function. Using an integral form of the Beta function the summation becomes
\begin{align}
S &= \sum_{n=2}^{\infty} \frac{1}{n \, (n+1)^{2} \, (n+2)} \\
&= \frac{1}{2} \, \int_{0}^{1} \left( \sum_{n=2}^{\infty} \frac{x^{n-1}}{n+1} \right) \, (1-x)^{2} \, dx \\
&= - \frac{1}{2} \, \int_{0}^{1} \left(x + \frac{x^{2}}{2} + \ln(1-x) \right) \, \frac{(1-x)^{2}}{x^{2}} \, dx \\
&= - \frac{1}{4} \, \int_{0}^{1} (2+x) (1-x)^{2} \, \frac{dx}{x} - \frac{1}{2} \, \int_{0}^{1} \left( 1 - \frac{1}{x} \right)^{2} \, \ln(1-x) \, dx \\
&= - \frac{1}{4} \left[ \frac{x^{3}}{3} - 3 x + 2 \ln(x) \right]_{0}^{1} - \frac{1}{2} \left[ (x-1) \, \ln(1-x) - x \right]_{0}^{1} + \left[ - Li_{2}(x) \right]_{0}^{1} \\ 
& \hspace{20mm} - \frac{1}{2} \left[ \frac{(x-1) \ln(1-x) - x \ln(x)}{x} \right]_{0}^{1} \\
&= \frac{2}{3} + \frac{1}{2} - Li_{2}(1) - \frac{1}{2} \, \lim_{x \to 0} \left\{ - \ln(x) - \frac{\ln(1-x)}{x} + \ln(x) \right\} \\
&= \frac{5}{3} - \zeta(2).  
\end{align}
This leads to the known result
\begin{align}
\sum_{n=2}^{\infty} \frac{1}{n \, (n+1)^{2} \, (n+2)} = \frac{5}{3} - \frac{\pi^{2}}{6}.
\end{align}
