Evaluating $\sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)}$ 
Show that series $$\sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)}$$ converges by simplifying its sequence of partial sums and find its sum.

I don't have much detail but this all I have:
$$\begin{align}\sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)}
&=\lim_{m\to\infty}\sum_{n=1}^{m}\frac{6}{m(m+1)(m+2)}\\
&=\lim_{m\to\infty}\frac{3(m^2+3m)}{2(m+1)(m+2)}\\
&=\frac{3}{2}\\
\end{align}$$
I know, I don't have much but any help will do with the detail or is it right.

Since 
\begin{gather*}
\frac{1}{n(n+1)(n+2)}=\frac{1}{2}\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right),
\end{gather*}
we have, by telescoping summation, 
\begin{align*}
\sum_{n=1}^N\frac{1}{n(n+1)(n+2)}&=\frac{1}{2}\sum_{n=1}^N\left(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)}\right)\\
&=\frac{1}{2}\sum_{n=1}^N\frac{1}{n(n+1)}-\frac{1}{2}\sum_{n=1}^N\frac{1}{(n+1)(n+2)}\\
&=\frac{1}{2}\sum_{n=1}^N\frac{1}{n(n+1)}-\frac{1}{2}\sum_{m=2}^{N+1}\frac{1}{m(m+1)}\quad \quad (m=n+1)\\
&=\frac{1}{2}\sum_{n=1}^N\frac{1}{n(n+1)}-\frac{1}{2}\sum_{n=2}^{N+1}\frac{1}{n(n+1)}\\
&=\frac{1}{2}\cdot\frac{1}{2}+\sum_{n=2}^N\frac{1}{n(n+1)}-\frac{1}{2}\sum_{n=2}^{N}\frac{1}{n(n+1)}-\frac{1}{2}\cdot\frac{1}{(N+1)(N+2)}\\
&=\frac{1}{2}\left(\frac{1}{2}-\frac{1}{(N+1)(N+2)}\right)\to \frac{1}{4}, \qquad \text{as } N\to\infty,
\end{align*}
from which we can infer that
\begin{gather*}
\sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)}=6\cdot\frac{1}{4}=3/2.
\end{gather*}
