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?
 A: 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...
A: $\begin{aligned}  \sum_{n \ge 1} \frac{1}{n(n+1)(n+2)}  &= \sum_{n \ge 1} \int_0^1\int_0^1\int_0^1 x^{n-1}y^nz^{n+1}\;{dx}\;{dy}\;{dz} \\& = \int_0^1\int_0^1\int_0^1\sum_{n \ge 1}  x^{n-1}y^nz^{n+1}\;{dx}\;{dy}\;{dz} \\
& = \int_0^1\int_0^1\int_0^1 \frac{z}{x}\sum_{n \ge 1} (xyz)^n\;{dx}\;{dy}\;{dz} \\& = \int_0^1\int_0^1\int_0^1\ \frac{yz^2}{1-xyz}\;{dx}\;{dy}\;{dz} \\
& = -\int_{0}^{1} \int_{0}^{1} z\log(1-zy)\;{dy}\;{dz} \\& = \int_{0}^{1} z+(1-z)\log(1-z)\;{dz}  \\& = \frac{1}{4}. \end{aligned} $
A: Partial fractions.
$$
n^3+3n^2+2n=n(n+1)(n+2),
$$
and thus
$$
\frac{1}{n^3+3n^2+2n} = \frac{A}{n}+\frac{B}{n+1}+\frac{C}{n+2}.
$$
You can determine the numbers $A,B,C$ by comparing coefficients of powers of $n$.
If you do that you may find that most terms in your sum cancel.
A: $$\frac{1}{n(n+1)(n+2)}=\frac{1}{n+2} \frac{1}{n(n+1)}=\frac{1}{n+2}\left( \frac{1}{n}- \frac{1}{n+1} \right)
=\frac{1}{n(n+2)}-\frac{1}{(n+1)(n+2)}=\frac{1}{2}\frac{1}{n}-\frac{1}{n+1}+\frac{1}{n+2}-\frac{1}{n+1}$$
A: We can avoid partial fraction and set up a telesoping series.
$\dfrac{1}{n(n+1)}=\dfrac{1}{n}-\dfrac{1}{n+1}$
Similarly, we have:
$$\dfrac{1}{n(n+1)}-\dfrac{1}{(n+1)(n+2)}=\dfrac{2}{n(n+1)(n+2)}$$
Note, that this is immediately apparent if you've seen some finite calculus.
A: $\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: $\displaystyle{\frac{1}{2\times1\times(1+1)} - \lim_{n\rightarrow +\infty}{\frac{1}{2n(n+1)}}} = \frac{1}{4}$
