# The irrationality of e

This is the post excerpt.

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.

Now,we proceed to finding an upper bound or A,closing in on our contradiction. $A<\sum_{k=1}^{\infty} \frac{1}{(q+1)^k}$

The infinite sum which provides the upper bound can be maximized by minimizing the value of q.Taking the minimum value of q which is 3. $A<(\sum_{k=1}^{\infty} \frac{1}{(4)^k}=\frac{1}{3}$

Since A is clearly non-negative,this is a contradiction as A must be an integer. $e$ isn’t rational.