<s>[INST] 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.
 [/INST] 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.
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