Sum of Sequence of Products of Consecutive Reciprocals/Proof 2

Theorem

$\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1}$

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:

$\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} }$

$=$

$\ds \sum_{j \mathop = 1}^n \paren {\frac 1 j - \frac 1 {j + 1} }$

$\ds $

$=$

$\ds 1 - \frac 1 {n + 1}$

$\ds $

$=$

$\ds \frac n {n + 1}$

$\blacksquare$