Sum of Sequence of Products of Consecutive Reciprocals/Proof 1

Theorem

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

Proof

Proof by induction:

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

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

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:

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

$\blacksquare$