# e  is irrational

Suppose e to be rational. Then there is a representation $e=\frac{n}{m}$ with $n,m\in {ℕ}^{\ast }$. For the integer

$\left(e-\sum _{i=0}^{m}\frac{1}{i!}\right)m!=\frac{n\cdot m!}{m}-\sum _{i=0}^{m}\frac{m!}{i!}=n\left(m-1\right)!-\sum _{i=0}^{m}\left(i+1\right)\cdot \left(i+2\right)\cdot \dots \cdot m\in ℤ$       [+]

we calculate the following estimate using the limit of the geometric series:

$\begin{array}{ll}0<\left(e-\sum _{i=0}^{m}\frac{1}{i!}\right)m!\hfill & =\sum _{i=0}^{\infty }\frac{m!}{i!}-\sum _{i=0}^{m}\frac{m!}{i!}\hfill \\ \hfill & =\sum _{i=m+1}^{\infty }\frac{m!}{i!}\hfill \\ \hfill & =\sum _{i=m+1}^{\infty }\frac{1}{\left(m+1\right)\cdot \left(m+2\right)\cdot \dots \cdot i}\hfill \\ \hfill & \le \sum _{i=m+1}^{\infty }\frac{1}{{\left(m+1\right)}^{i-m}}\hfill \\ \hfill & =\sum _{i=0}^{\infty }\frac{1}{{\left(m+1\right)}^{i+1}}\hfill \\ \hfill & =\frac{1}{m+1}\cdot \frac{1}{1-\frac{1}{m+1}}=\frac{1}{m}<1\hfill \end{array}$

According to [+] this proves the existance of an integer within $\right]0,1\left[$.   Contradiction!