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 $$ \sum_{n=1}^N\frac{1}{n(n+1)(n+2)} =\frac{1}{2}-\frac{1}{2(N+1)}+\frac{1}{2(N+2)}-\frac{1}{4}. $$

Using Euler's beta function, $$S=\sum_{k\geq 1}\frac{(k-1)!}{(k+2)!}=\sum_{k\geq 1}\frac{\Gamma(k)}{\Gamma(k+3)}=\frac{1}{2}\sum_{k\geq 1}B(k,3)$$ hence: $$ S = \frac{1}{2}\int_{0}^{1}(1-x)^2\sum_{k\geq 1}x^{k-1}\,dx=\frac{1}{2}\int_{0}^{1}(1-x)\,dx = \color{red}{\frac{1}{4}}. $$ A straightforward generalization of this approach gives the identity: $$ \sum_{n\geq 1}\frac{1}{n(n+1)\cdots(n+N)}=\color{red}{\frac{1}{N\cdot N!}}.$$

There is an alternate method and is as follows.

Notice that $$ \frac{1}{n(n+1)(n+2)} = \frac{(n-1)!}{(n+2)!} = \frac{1}{2!} \, B(n,3) $$ where $B(x,y)$ is the Beta function. Using an integral form of the Beta function the summation becomes \begin{align} S &= \sum_{n=1}^{\infty} \frac{1}{n \, (n+1) \, (n+2)} \\ &= \frac{1}{2} \, \int_{0}^{1} \left( \sum_{n=1}^{\infty} x^{n-1} \right) \, (1-x)^{2} \, dx \\ &= \frac{1}{2} \, \int_{0}^{1} \frac{(1-x)^{2}}{1-x} \, dx = \frac{1}{2} \, \int_{0}^{1} (1-x) \, dx \\ &= \frac{1}{4} \end{align}

This leads to the known result \begin{align} \sum_{n=1}^{\infty} \frac{1}{n \, (n+1) \, (n+2)} = \frac{1}{4}. \end{align}