Sum of Sequence of Products of Consecutive Reciprocals/Proof 1

Proof
Proof by induction:

For all $n \in \N_{>0}$, let $P \left({n}\right)$ be the proposition:
 * $\displaystyle \forall n \ge 1: \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1}$

Basis for the Induction
$\map P 1$ 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 $\map P k$ is true, where $k \ge 1$, then it logically follows that $\map P {k + 1}$ is true.

So this is our induction hypothesis:
 * $\displaystyle \sum_{j \mathop = 1}^k \frac 1 {j \paren {j + 1} } = \frac k {k + 1}$

Then we need to show:
 * $\displaystyle \sum_{j \mathop = 1}^{k + 1} \frac 1 {j \paren {j + 1} } = \frac {k + 1} {k + 2}$

Induction Step
This is our induction step:

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

Therefore:
 * $\displaystyle \forall n \ge 1: \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = \frac n {n + 1}$