Evaluate $\sum_{n=1}^{\infty}\frac{1}{n^3+3 n^2+2 n}$ Summing this series from $0$ to $\infty$, the result is $\frac{1}{4}$. I tried a lot, but I could not get this result. I think it´s wrong.
Can anybody help me?

$$\frac{1}{n(n+1)(n+2)}=\frac{1}{n+2} \frac{1}{n(n+1)}=\frac{1}{n+2}\left( \frac{1}{n}- \frac{1}{n+1} \right)
=\frac{1}{n(n+2)}-\frac{1}{(n+1)(n+2)}=\frac{1}{2}\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}-\frac{1}{n+1}$$
