Sum of Sequence of Products of Consecutive Reciprocals/Proof 2

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

and that $\ds \sum_{j \mathop = 1}^n \paren {\frac 1 j - \frac 1 {j + 1} }$ is a telescoping series.

Therefore: