Prove that $\sum_{x=1}^{n} \frac{1}{x (x+1)(x+2)} = \frac{1}{4} - \frac{1}{2 (n+1) (n+2)}$ [closed]

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.

Solution 1:

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.

Solution 2:

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)} $