# Find the sum of $\sum_{r=1}^n\frac{1}{r(r+1)(r+2)}$ [duplicate]

This question is posing a few issues for me, so I've started by breaking the fraction down into partial fractions as usual but when i started to write out each of the terms they don't seem to cancel in any order or pattern that i can spot. any help would be appreciated.

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

## marked as duplicate by StubbornAtom, amWhy, Paul Frost, Leucippus, max_zornOct 4 '18 at 3:34

• Haha, sorry. Consider $\frac 12\left(\sum(\frac 1r-\frac 1{r+1}\right)+\frac 12\left(\sum(\frac 1{r+2}-\frac 1{r+1}\right)$ – AnotherJohnDoe Oct 3 '18 at 17:17
Note that$$(\forall r\in\mathbb{N}):\frac1{r(r+1)(r+2)}=\frac12\left(\frac1{r(r+1)}-\frac1{(r+1)(r+2)}\right).$$So, your series is a telescoping series.
A method using the Beta function is the following: \begin{align} \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)} &= \sum_{n=1}^{\infty} \frac{(n-1)!}{(n+2)!} \\ &= \sum_{n=0}^{\infty} \frac{\Gamma(n+1)}{\Gamma(n+4)} = \frac{1}{\Gamma(3)} \, \sum_{n=0}^{\infty} B(n+1,3) \\ &= \frac{1}{2} \, \sum_{n=0}^{\infty} \, \int_{0}^{1} t^{2} (1-t)^{n} \, dt \\ &= \frac{1}{2} \, \int_{0}^{1} \frac{t^{2} \, dt}{1-(1-t)} = \frac{1}{2} \, \int_{0}^{1} t \, dt \\ &= \frac{1}{4} \end{align}