Rewriting a simple fraction to fit the Harmonic series

Question

I'm refreshing my brain by going through some chapters of CLRS (algorithms bible). One of the first example problems in the appendix (A.1) is showing the harmonic series. But they do a simplification of the fraction that leaves me dumbfounded. I understand the sum, but not how they went from this fraction:

$$\frac{1}{x(x+1)}$$

to this one:

$$\frac{1}{x} - \frac{1}{x+1}$$

I can work it backwards fine, but starting from $\frac{1}{x(x+1)}$ I can't wrap my head around it! What am I forgetting?

The original problem was to find an equation for this harmonic series:

$$\sum_{x=1}^n \frac{1}{x(x+1)}$$

Answer 1

Write $$\frac{1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}$$ and solve for $A$ and $B$.

Answer 2

This generalizes nicely to the product of any number of consecutive integers.

If $p_n(x) =x(x+1)...(x+n-1) =\prod_{j=0}^{n-1} (x+j) $, you can get telescoping sums for both $p_n(x)$ and $\dfrac1{p_n(x)}$.

$\begin{array}\\ p_n(x+1)-p_n(x) &=\prod_{j=0}^{n-1} (x+1+j)-\prod_{j=0}^{n-1} (x+j)\\ &=\prod_{j=1}^{n} (x+j)-\prod_{j=0}^{n-1} (x+j)\\ &=\prod_{j=1}^{n-1} (x+j)((x+n)-x)\\ &=n\prod_{j=0}^{n-2} (x+1+j)\\ &=np_{n-1}(x+1) \end{array} $

and

$\begin{array}\\ \dfrac1{p_n(x)}-\dfrac1{p_n(x+1)} &=\dfrac1{\prod_{j=0}^{n-1} (x+j)}-\dfrac1{\prod_{j=0}^{n-1} (x+1+j)}\\ &=\dfrac1{\prod_{j=0}^{n-1} (x+j)}-\dfrac1{\prod_{j=1}^{n} (x+j)}\\ &=\dfrac1{\prod_{j=0}^{n} (x+j)}((x+n)-x)\\ &=\dfrac{n}{p_{n+1}(x)}\\ \end{array} $

This last, for $n=1$ and $2$ is

$\dfrac1{x}-\dfrac1{x+1} =\dfrac1{x(x+1)} $ and $\dfrac1{x(x+1)}-\dfrac1{(x+1)(x+2)} =\dfrac{2}{x(x+1)(x+2)} $.