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

Solution

1. The general term of the series can be decomposed using partial fractions as $\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)$.
2. Writing out the partial sums, we observe that the series telescopes: $$\begin{array}{rl}{S}_N& =\frac{1}{2}\sum _{n=1}^N\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\\ & =\frac{1}{2}\left(\frac{1}{1\cdot 2}-\frac{1}{(N+1)(N+2)}\right)\end{array}$$.
3. As $N\to \infty$, the term $\frac{1}{(N+1)(N+2)}$ approaches $0$, leaving the sum as $\frac{1}{2}\cdot \frac{1}{2}=\frac{1}{4}$.
4. Hence the answer is (B).