Find the sum of the series $\sum \frac{1}{n(n+1)(n+2)}$

Hint

$$\frac{2}{n(n+2)}=\frac{1}{n}-\frac{1}{n+2}$$

Now multiply both sides by $\frac{1}{n+1}$.

Using Partial Fraction Decomposition, $$\frac1{n(n+1)(n+2)}=\frac An+\frac B{n+1}+\frac C{n+2}$$

$$\implies 1=A(n+1)(n+2)+Bn(n+2)+Cn(n+1)$$ $$\implies 1=n^2(A+B+C)+n(3A+2B+C)+2A$$

Comparing the coefficients of the different powers (namely, $0,1,2$) of $n,$ we get $A=\frac12,B=-1,C=\frac12$

$$\implies\frac1{n(n+1)(n+2)}=\frac12\cdot\frac1n-\frac1{n+1}+\frac12\cdot\frac1{n+2}$$ $$=-\frac12\left(\underbrace{\frac1{n+1}-\frac1n}\right)+\frac12\left(\underbrace{\frac1{n+2}-\frac1{n+1}}\right)$$

Can you recognize the two Telescoping series?

Hint. You may write $$\frac{1}{n(n+1)(n+2)} =\frac{1}{2}\left(\frac{1}{n}-\frac{1}{(n+1)}\right)+\frac{1}{2}\left(\frac{1}{(n+2)}-\frac{1}{(n+1)}\right) $$ giving two telescoping sums $$ \sum_{n=1}^N\frac{1}{n(n+1)(n+2)} =\frac{1}{2}-\frac{1}{2(N+1)}+\frac{1}{2(N+2)}-\frac{1}{4}. $$