Test the condition for convergence of \sum_{n=1}^\infty\frac{1}{n(n+1)(n+2)} and find the sum if

There is an alternate method and is as follows.
Notice that
${\displaystyle \frac{1}{n(n+1)(n+2)}=\frac{(n-1)!}{(n+2)!}=\frac{12}{B}(n,3)}$
where ${\displaystyle B(x,y)}$ is the Beta function. Using an integral form of the Beta function the summation becomes
${\displaystyle S=\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)(n+2\}}}$
$=\frac{1}{2}{\int }_0^1(\sum _{n=1}^{\mathrm{\infty }}{x}^{n-1})(1-x{)}^{2}dx$
$=\frac{1}{2}{\int }_0^1\frac{(1-x{)}^2}{1-x}dx=\frac{1}{2}{\int }_0^1(1-x)dx$
${\displaystyle =\frac{1}{4}}$
This leads to the known result
${\displaystyle \sum _{n=1}^{\mathrm{\infty }}\frac{1}{n(n+1)(n+2)}=\frac{1}{4}}$