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).

You may first write a partial fraction decomposition, giving
$$
\frac{1}{n(n+1)^2(n+2)}=\frac12\left(\frac{1}n-\frac1{n+2}\right)-\frac{1}{(1+n)^2}
$$ then obtain your sum, by telescoping for the first two terms above, then by identifying a celebrated series for the third term. You get
$$
\sum_2^N\frac{1}{n(n+1)^2(n+2)}=\frac{5}{12}-\frac{1}{2 (1+N)}-\frac{1}{2 (2+N)}-\sum_2^N\frac{1}{(n+1)^2}
$$ and, as $N \to \infty$,

$$
\sum_{n=2}^\infty\frac{1}{n(n+1)^2(n+2)}=\frac53-\frac{\pi ^2}{6}=\color{red}{0.021732\cdots}.
$$ 

