I would like to calculate the sum of the following series by modifying the fraction.
$\sum_{n=1}^{\infty}\frac{1}{n(n+1)(n+2)}$
I came across an explanation that this fraction is equal to $\frac{1}{2}(\frac{1}{n(n+1)}-\frac{1}{(n+1)(n+2)})$
It is easy then to calculate the sum which is equal to $\frac{1}{4}$, but I don't understand how to come up with that modification of a fraction. I would very much appreciate your help.
On
Here is a simple way to get the desired expression:
Take some reals $x \neq 0$ and $y \neq -x$, so that the fractions $\frac{1}{x}$ and $\frac{1}{x+y}$ are defined.
Then: $$\frac{1}{x} - \frac{1}{x+y} = \frac{x + y - x}{x(x+y)} = \frac{y}{x(x+y)}$$
As such, if $y$ is also different from $0$: $$\frac{1}{x(x+y)} = \frac{1}{y} \left(\frac{1}{x} - \frac{1}{x+y}\right)$$
This gives you here: $$\frac{1}{n(n+2)} = \frac{1}{2} \left(\frac{1}{n} - \frac{1}{n+2}\right)$$ and then you can just multiply by $\frac{1}{n+1}$.
Here is a way to do the fraction like you posted: $$\begin{array}{l} \displaystyle\frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{a}{n} + \frac{b}{{n + 1}} + \frac{c}{{n + 2}}\\ \Leftrightarrow a\left( {n + 1} \right)\left( {n + 2} \right) + bn\left( {n + 2} \right) + cn\left( {n + 1} \right) = 1\\ \displaystyle\Rightarrow \left\{ \begin{array}{l} a + b + c = 0\\ 3a + 2b + c = 0\\ 2a = 1 \end{array} \right. \displaystyle\Rightarrow \left\{ \begin{array}{l} a = \frac{1}{2}\\ b = - 1\\ c = \frac{1}{2} \end{array} \right. \displaystyle\Rightarrow \frac{1}{{n\left( {n + 1} \right)\left( {n + 2} \right)}} = \frac{1}{{2n}} - \frac{1}{{n + 1}} + \frac{1}{{2\left( {n + 2} \right)}}\\ \displaystyle\frac{1}{{2n}} - \frac{1}{{n + 1}} + \frac{1}{{2\left( {n + 2} \right)}} = \frac{1}{{2n}} - \frac{1}{{2\left( {n + 1} \right)}} - \left( {\frac{1}{{2\left( {n + 1} \right)}} - \frac{1}{{2\left( {n + 2} \right)}}} \right) = \frac{1}{2}\left( {\frac{1}{{n\left( {n + 1} \right)}} - \frac{1}{{\left( {n + 1} \right)\left( {n + 2} \right)}}} \right) \end{array}$$