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

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## 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 $\map P n$ 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:

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

\(\displaystyle \) | \(=\) | \(\displaystyle \frac k {k + 1} + \frac 1 {\paren {k + 1} \paren {k + 2} }\) | Induction hypothesis | ||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {k \paren {k + 2} + 1} {\paren {k + 1} \paren {k + 2} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {k^2 + 2 k + 1} {\paren {k + 1} \paren {k + 2} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {\paren {k + 1}^2} {\paren {k + 1} \paren {k + 2} }\) | |||||||||||

\(\displaystyle \) | \(=\) | \(\displaystyle \frac {k + 1} {k + 2}\) |

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

Therefore:

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

$\blacksquare$