e is irrational

Suppose e to be rational. Then there is a representation

e = \frac{n}{m}

with

n, m \in ℕ^{*}

. For the integer

(e - \sum_{i = 0}^{m} \frac{1}{i!}) m! = \frac{n \cdot m!}{m} - \sum_{i = 0}^{m} \frac{m!}{i!} = n (m - 1)! - \sum_{i = 0}^{m} (i + 1) \cdot (i + 2) \cdot \dots \cdot m \in ℤ

[+]

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

\begin{array}{l} 0 < (e - \sum_{i = 0}^{m} \frac{1}{i!}) m! & = \sum_{i = 0}^{\infty} \frac{m!}{i!} - \sum_{i = 0}^{m} \frac{m!}{i!} \\ = \sum_{i = m + 1}^{\infty} \frac{m!}{i!} \\ = \sum_{i = m + 1}^{\infty} \frac{1}{(m + 1) \cdot (m + 2) \cdot \dots \cdot i} \\ \leq \sum_{i = m + 1}^{\infty} \frac{1}{{(m + 1)}^{i - m}} \\ = \sum_{i = 0}^{\infty} \frac{1}{{(m + 1)}^{i + 1}} \\ = \frac{1}{m + 1} \cdot \frac{1}{1 - \frac{1}{m + 1}} = \frac{1}{m} < 1 \end{array}

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