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.
 A: $$\begin{align}
\sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)}
&=\sum_{n=1}^{\infty}\left(\frac{3}{n}+\frac{3}{n+2}-\frac{6}{n+1}\right)\tag{1}\\
&=\sum_{n=1}^{\infty}\int_{0}^{1}\left(3x^{n-1}+3x^{n+1}-{6x}^{n}\right)\,\mathrm dx\tag{2}\\
&=\int_{0}^{1}\sum_{n=1}^{\infty}\left(3x^{n-1}+3x^{n+1}-{6x}^{n}\right)\,\mathrm dx\tag{3}\\
&=\int_{0}^{1}\frac{3}{1-x}+\frac{3x^2}{1-x}-\frac{6x}{1-x}\,\mathrm dx\tag{4}\\
&=\int_{0}^{1}\left(\frac{3+3x^2-6x}{1-x}\right)\,\mathrm dx\tag{5}\\
&=3\int_{0}^{1}\frac{x^2-2x+1}{1-x}\,\mathrm dx\tag{6}\\
&=3\int_{0}^{1}{1-x}\,\mathrm dx\tag{7}\\
&=3\cdot\frac14\\
&=\frac32\\
\end{align}$$

$$\large\sum_{n=1}^{\infty}\frac{6}{n(n+1)(n+2)}=\frac32\\$$

A: 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*}
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$$
\sum_{n\ =\ 1}^{\infty}x^{n - 1}={1 \over 1 -x}\ \imp\
\sum_{n\ =\ 1}^{\infty}{x^{n} \over n}=\int_{0}^{x}{\dd t \over 1 - t}
=-\ln\pars{1 -x}
$$

$$
\sum_{n\ =\ 1}^{\infty}{x^{n + 1} \over n\pars{n + 1}}
=-\int_{0}^{x}\ln\pars{1 - t}\,\dd t 
=x + \pars{1 - x}\ln\pars{1 - x}
$$

$$
\sum_{n\ =\ 1}^{\infty}{1 \over n\pars{n + 1}\pars{n + 2}}
=\int_{0}^{1}\bracks{t + \pars{1 - t}\ln\pars{1 - t}}\,\dd t 
={1 \over 4}
$$

$$\color{#66f}{\large%
\sum_{n\ =\ 1}^{\infty}{6 \over n\pars{n + 1}\pars{n + 2}}} 
=6\times{1 \over 4}=\color{#66f}{\large{3 \over 2}}
$$

