Sum of Sequence of Products of Consecutive Reciprocals

Theorem

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

Proof
Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \forall n \ge 1: \sum_{j = 1}^n \frac 1 {j \left({j+1}\right)} = \frac n {n+1}$

Basis for the Induction

 * $P(1)$ is true, as this just says $\dfrac 1 2 = \dfrac 1 2$.

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $P \left({k}\right)$ is true, where $k \ge 1$, then it logically follows that $P \left({k+1}\right)$ is true.

So this is our induction hypothesis:
 * $\displaystyle \sum_{j = 1}^k \frac 1 {j \left({j+1}\right)} = \frac k {k+1}$

Then we need to show:
 * $\displaystyle \sum_{j = 1}^{k+1} \frac 1 {j \left({j+1}\right)} = \frac {k+1} {k+2}$

Induction Step
This is our induction step:

So $P \left({k}\right) \implies P \left({k+1}\right)$ and the result follows by the Principle of Mathematical Induction.

Therefore:
 * $\displaystyle \forall n \ge 1: \sum_{j = 1}^n \frac 1 {j \left({j+1}\right)} = \frac n {n+1}$

Proof By Telescoping Sums
We can observe that $\displaystyle \frac j {j+1} = \frac 1 {j} - \frac 1 {j+1}$ and that $\displaystyle\sum_{j = 1}^n \left({\frac 1 {j} - \frac 1 {j+1}}\right)$ is a telescoping sum.

Therefore: