$\displaystyle \sum_{n=1}^{\infty} \frac{1}{n(n+1)(n+2)}$
$\displaystyle \frac{u_{n+1}}{u_n}=\ldots=\frac{n}{n+3}=\frac{ \alpha n+\beta}{\alpha n+\gamma}$ (hipergeometric series) . According to a well known theorem , $\displaystyle S=\frac{u_1\gamma}{\gamma-(\alpha+\beta)}=\ldots=\frac{1}{4}$
Note that
$\displaystyle \frac{1}{n(n+1)(n+2)} = \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+1}\right) - \frac{1}{2}\left(\frac{1}{n+1} - \frac{1}{n+2}\right)$
Write some terms in the series
$\displaystyle \sum_{n=1}^{\infty} \frac{1}{2}\left(\frac{1}{n} - \frac{1}{n+1}\right) - \frac{1}{2}\left(\frac{1}{n+1} - \frac{1}{n+2}\right)$
and see what happens.