# Sum of Sequence of Products of Consecutive Reciprocals/Proof 1

## Theorem

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

## Proof

Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:

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

### 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:

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

Then we need to show:

$\displaystyle \sum_{j \mathop = 1}^{k + 1} \frac 1 {j \left({j + 1}\right)} = \frac {k + 1} {k + 2}$

### Induction Step

This is our induction step:

 $\displaystyle \sum_{j \mathop = 1}^{k + 1} \frac 1 {j \left({j + 1}\right)}$ $=$ $\displaystyle \sum_{j \mathop = 1}^k \frac 1 {j \left({j + 1}\right)} + \frac 1 {\left({k + 1}\right) \left({k + 2}\right)}$ $\displaystyle$ $=$ $\displaystyle \frac k {k + 1} + \frac 1 {\left({k + 1}\right) \left({k + 2}\right)}$ Induction hypothesis $\displaystyle$ $=$ $\displaystyle \frac {k \left({k + 2}\right) + 1} {\left({k + 1}\right) \left({k + 2}\right)}$ $\displaystyle$ $=$ $\displaystyle \frac {k^2 + 2 k + 1} {\left({k + 1}\right) \left({k + 2}\right)}$ $\displaystyle$ $=$ $\displaystyle \frac {\left({k + 1}\right)^2} {\left({k + 1}\right) \left({k + 2}\right)}$ $\displaystyle$ $=$ $\displaystyle \frac {k + 1} {k + 2}$

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

Therefore:

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

$\blacksquare$