Sum of Sequence of Products of Consecutive Reciprocals

Theorem

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

Proof
Proof by induction:

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

Basis for the Induction

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

This is our basis for the induction.

Induction Hypothesis

 * Now we need to show that, if $$P \left({k}\right)$$ is true, where $$k \ge 2$$, then it logically follows that $$P \left({k+1}\right)$$ is true.

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

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

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