The sum of the infinite series ${\sum }_{n=1}^{\infty }\frac{1}{n(n+1)(n+2)}$ is:

Solution

1. The general term is $$\begin{array}{rl}{a}_n& =\frac{1}{n(n+1)(n+2)}\\ & =\frac{1}{2}[\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}]\end{array}$$.
2. The sum is a telescoping series $S=\frac{1}{2}[(\frac{1}{1\ast 2}-\frac{1}{2\ast 3})+(\frac{1}{2\ast 3}-\frac{1}{3\ast 4})+...]$.
3. All intermediate terms cancel out, leaving $$\begin{array}{rl}S& =\frac{1}{2}\ast \frac{1}{2}\\ & =\frac{1}{4}\end{array}$$.
4. Hence the answer is (B).