Prove that $\sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$ 
Prove that $\displaystyle \sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$.

I tried using the partial fraction decomposition $a_j = \frac{1}{2j} - \frac{1}{j+1} + \frac{1}{2(j+2)}$, but I don't see how that helps.
 A: Hint. From what you wrote one may observe that
$$
\frac{1}{j (j+1)(j+2)}=\frac{1}{2j} - \frac{1}{j+1} + \frac{1}{2(j+2)}=\frac12\left(\frac1j - \frac1{j+1}\right)+ \frac12\left(\frac1{j+2} - \frac1{j+1}\right)
$$ then noticing that terms telescope.

Remark. One may also write
$$
\frac{1}{j (j+1)(j+2)}=\frac12\frac1{j(j+1)}-\frac12\frac1{(j+1)(j+2)}
$$ then by summing terms telescope giving

$$
\sum_{j=1}^n\frac{1}{j (j+1)(j+2)}=\frac12\frac1{1\times(1+1)}-\frac12\frac1{(n+1)(n+2)}= \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}
$$ 

as announced.
A: More generally,
if
$x_{(n)}
=\prod_{k=0}^{n-1} (x+k)
$,
$\begin{array}\\
\dfrac1{x_{(n)}}-\dfrac1{(x+1)_{(n)}}
&=\dfrac1{\prod_{k=0}^{n-1} (x+k)}-\dfrac1{\prod_{k=0}^{n-1} (x+1+k)}\\
&=\dfrac1{\prod_{k=0}^{n-1} (x+k)}-\dfrac1{\prod_{k=1}^{n} (x+k)}\\
&=\dfrac1{\prod_{k=1}^{n-1} (x+k)}\left(\dfrac1{x}-\dfrac1{x+n}\right)\\
&=\dfrac1{\prod_{k=1}^{n-1} (x+k)}\left(\dfrac{n}{x(x+n)}\right)\\
&=\dfrac{n}{\prod_{k=0}^{n} (x+k)}\\
&=\dfrac{n}{x_{(n+1)}}\\
\end{array}
$
Therefore
$\begin{array}\\
\sum_{j=1}^m \dfrac{1}{j_{(n+1)}}
&=\sum_{j=1}^m \dfrac1{n}\left(\dfrac1{j_{(n)}}-\dfrac1{(j+1)_{(n)}}\right)\\
&= \dfrac1{n}\left(\dfrac1{1_{(n)}}-\dfrac1{(m+1)_{(n)}}\right)\\
\text{or}\\
\sum_{j=1}^m \dfrac{1}{\prod_{k=0}^{n} (j+k)}
&= \dfrac1{n}\left(\dfrac1{n!}-\dfrac1{\prod_{k=0}^{n-1} (m+1+k)}\right)\\
\end{array}
$
If
$n=2$,
this becomes
$\sum_{j=1}^m \dfrac{1}{\prod_{k=0}^{2} (j+k)}
= \dfrac1{2}\left(\dfrac1{2!}-\dfrac1{\prod_{k=0}^{1} (m+1+k)}\right)
$
or
$\sum_{j=1}^m \dfrac{1}{j(j+1)j+2)}
= \dfrac1{4}-\dfrac1{2(m+1)(m+2)}
$
A: Note that we can write
$$\begin{align}
\frac{1}{m(m+1)(m+2)}&=\frac12\left(\frac{(m+2)-m}{m(m+1)(m+2)}\right)\\\\
&=\frac12\left(\frac{1}{m(m+1)}-\frac{1}{(m+1)(m+2)}\right)
\end{align}$$
So, letting $a_m=\frac{1}{m(m+1)}$, we see that
$$\begin{align}
\frac{1}{m(m+1)(m+2)}&=\frac12\left(a_m-a_{m+1}\right)
\end{align}$$
from which we find
$$\begin{align}
\sum_{m=1}^n \frac{1}{m(m+1)(m+2)}&=\sum_{m=1}^n \frac12\left(a_m-a_{m+1}\right)\\\\
&=\frac12(a_1-a_{n+1})\\\\
&=\frac14-\frac12 \frac{1}{(n+1)(n+2)}
\end{align}$$
as was to be shown!
