# Sum of Sequence of Products of Consecutive Reciprocals

## Theorem

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

### Corollary

$\ds \lim_{n \mathop \to \infty} \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} } = 1$

## Proof 1

Proof by induction:

For all $n \in \N_{>0}$, let $\map P n$ be the proposition:

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

### 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:

$\ds \sum_{j \mathop = 1}^k \frac 1 {j \paren {j + 1} } = \frac k {k + 1}$

Then we need to show:

$\ds \sum_{j \mathop = 1}^{k + 1} \frac 1 {j \paren {j + 1} } = \frac {k + 1} {k + 2}$

### Induction Step

This is our induction step:

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

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

Therefore:

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

$\blacksquare$

## Proof 2

We can observe that:

$\dfrac 1 {j \paren {j + 1} } = \dfrac 1 j - \dfrac 1 {j + 1}$

and that $\ds \sum_{j \mathop = 1}^n \paren {\frac 1 j - \frac 1 {j + 1} }$ is a telescoping series.

Therefore:

 $\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} }$ $=$ $\ds \sum_{j \mathop = 1}^n \paren {\frac 1 j - \frac 1 {j + 1} }$ Telescoping Series: Example 1 $\ds$ $=$ $\ds 1 - \frac 1 {n + 1}$ $\ds$ $=$ $\ds \frac n {n + 1}$

$\blacksquare$

## Proof 3

Observe that:

 $\ds \int_j^{j + 1} {\dfrac {\rd x} {x^2} }$ $=$ $\ds \intlimits {\dfrac {-1} x} j {j + 1}$ Primitive of Power $\ds$ $=$ $\ds \dfrac 1 j - \dfrac 1 {j + 1}$ $\ds$ $=$ $\ds \dfrac 1 {j \paren {j + 1} }$

Therefore:

 $\ds \sum_{j \mathop = 1}^n \frac 1 {j \paren {j + 1} }$ $=$ $\ds \sum_{j \mathop = 1}^n \int_j^{j + 1} {\dfrac {\rd x} {x^2} }$ $\ds$ $=$ $\ds \int_1^{n + 1} {\dfrac {\rd x} {x^2} }$ $\ds$ $=$ $\ds \intlimits {\dfrac {-1} x} 1 {n + 1}$ Primitive of Power $\ds$ $=$ $\ds \frac n {n + 1}$

$\blacksquare$