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

Question

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.

Answer 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.

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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.

Answer 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)} $

Answer 3

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!