Telescoping series with partial fractions

First series. $\frac1{n(n+2)}=\frac12\left(\frac1n-\frac1{n+2}\right)$. Then
$$S_N=\frac12\sum_{n=1}^N\left(\frac1n-\frac1{n+2}\right)=\frac12\left(1+\frac12-\frac1{N+1}-\frac1{N+2}\right)\longrightarrow\frac34.$$
The terms cancel *two* steps apart, so two terms of each end survive.

Second series. $\frac1{n(n+1)(n+2)}=\frac12\left(\frac1{n(n+1)}-\frac1{(n+1)(n+2)}\right)$. The sum telescopes:
$$S_N=\frac12\left(\frac1{1\cdot2}-\frac1{(N+1)(N+2)}\right)\longrightarrow\frac14.$$