$\sum_1^\infty{\frac{1}{n(n+1)(n+2)}}$?

Question

How to find the sum of the series $\sum_1^\infty{\frac{1}{n(n+1)(n+2)}}$?

I expanded it via partial fractions but it does not look like a telescoping series which I was expecting.

Am I missing something obvious or easy manipulation here?

$$\frac{1}{n(n+1)(n+2)} = \frac{1}{2n}-\frac{1}{2(n+1)}+\frac{1}{2(n+2)}-\frac{1}{2(n+1)}$$ — Gabriel Romon, Jun 15 '17 at 06:05
... telescoping series $;\frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left(\frac{1}{n (n+1)} - \frac{1}{(n+1)(n+2)}\right)$ — dxiv, Jun 15 '17 at 06:05
See also: https://math.stackexchange.com/questions/560816/find-the-sum-of-the-series-sum-frac1nn1n2/560897#560897 — lab bhattacharjee, Jun 15 '17 at 06:07
take limit to finite representation of the sum, which is $\tfrac{1}{4}-\tfrac{1}{2(n+1)(n+2)}$ — serg_1, Jun 15 '17 at 06:50

score 4 · Answer 1 · answered Jun 15 '17 at 06:05

4

HINT:

$$\dfrac2{n(n+1)(n+2)}=\dfrac{n+2-n}{n(n+1)(n+2)}=\dfrac1{n(n+1)}-\dfrac1{(n+1)(n+2)}=f(n)-f(n+1)$$ where $f(m)=\dfrac1{m(m+1)}$

answered Jun 15 '17 at 06:05

lab bhattacharjee

1 Answers1