The irrationality of e

The irrationality of e can be proved using the infinite series expansion.

$e=\sum_{k=0}^{\infty} \frac{1}{k!}$

Assume that e is rational.i.e $e=\frac{p}{q}$ where $p \in \mathbb{Z}$ and $q \in \mathbb{Z}-{0}$ .

Since e is positive and 2.5<e<3(I’m not going to prove this.It’s quite simple. $\frac{5}{2}=1+1+\frac{1}{2}$ ,use this as a lower bound for e and proceed to the upper bound)

This ensures that $q \neq 2$ ie $q \geq 3$

$e=\frac{p}{q}=\sum_{k=0}^{\infty} \frac{1}{k!} =\sum_{k=0}^{q} \frac{1}{k!}+\frac{1}{(q+1)!}+\frac{1}{(q+2)!}+\cdots$

Multiplying both sides by $q!$ ;

$p(q-1)!=q!(\sum_{k=0}^{q} \frac{1}{k!})+(\frac{1}{q+1}+\frac{1}{(q+1)(q+2)}+\frac{1}{(q+1)(q+2)(q+3)}+\cdots)$

$p(q-1)!$ and the sum $q!(\sum_{k=0}^{q} \frac{1}{k!})$ are integers.So the infinite sum $A=\frac{1}{q+1}+\frac{1}{(q+1)(q+2)}+\frac{1}{(q+1)(q+2)(q+3)}+\cdots$ must also be an integer.