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}{2(n+1)}(\frac{1}{n}-\frac{1}{n+2}) = $
$ = \frac{1}{2n(n+1)} - \frac{1}{2(n+1)(n+2)}$
So the sum equals: $\displaystyle{\frac{1}{2\times1\times(1+1)} - \lim_{n\rightarrow +\infty}{\frac{1}{2n(n+1)}}} = \frac{1}{4}$
