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?

Hints - Partial Fractions:
$$n^3+3n^2+2n=n(n+1)(n+2)\Longrightarrow$$
$$\frac{1}{n^3+3n^2+2n}=\frac{1}{n(n+1)(n+2)}=\frac{1}{2n}-\frac{1}{n+1}+\frac{1}{2(n+2)}\Longrightarrow$$
$$\sum_{n=1}^\infty\frac{1}{n^3+3n^2+2n}=\left(\frac{1}{2}-\frac{1}{2}+\frac{1}{6}\right)+\left(\frac{1}{4}-\frac{1}{3}+\frac{1}{8}\right)=\ldots$$
Further hint: find the cancelation pattern in the above telescopic-ish series...
