Sum of Sequence of Products of Consecutive Reciprocals/Proof 2

Theorem

 * $\displaystyle \sum_{j \mathop = 1}^n \frac 1 {j \left({j+1}\right)} = \frac n {n+1}$

Proof
We can observe that:
 * $\displaystyle \frac j {j+1} = \frac 1 j - \frac 1 {j+1}$

and that $\displaystyle \sum_{j \mathop = 1}^n \left({\frac 1 j - \frac 1 {j+1}}\right)$ is a Telescoping Series.

Therefore: