Sum of Sequence of Products of Consecutive Reciprocals

Theorem

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

Proof

Proof by induction:

For all $n \in \N^*$, let $P \left({n}\right)$ be the proposition:

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

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:

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

Then we need to show:

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

Induction Step

This is our induction step:

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

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

Therefore:

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

$\blacksquare$

Proof By Telescoping Sums

We can observe that $\displaystyle \frac j {j+1} = \frac 1 {j} - \frac 1 {j+1}$ and that $\displaystyle\sum_{j = 1}^n \left({\frac 1 {j} - \frac 1 {j+1}}\right)$ is a telescoping sum.

Therefore:

	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \sum_{j = 1}^n \frac 1 {j \left({j+1}\right)}\)	\(=\)	\(\displaystyle \sum_{j = 1}^n \left({\frac 1 {j} - \frac 1 {j+1} }\right)\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle 1 - \frac 1 {n+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)
	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)	\(=\)	\(\displaystyle \frac {n} {n+1}\)	\(\displaystyle \)	\(\displaystyle \)	\(\displaystyle \)

$\blacksquare$

Sources

• George E. Andrews: Number Theory (1971): $\S 1.1$: Exercise $6$
• Gary Chartrand: Introductory Graph Theory (1977): Appendix $\text{A}.6$: Problem $39$