The Euler number e is usually defined as:
(1)$$ \begin{align} e = \sum_{n=0}^\infty \frac{1}{n!} \end{align} $$
Let sn be the sequence of partial sums:
(2)$$ \begin{align} s_n = \sum_{k=0}^n \frac{1}{k!} \end{align} $$
The partial sums sn converge rapidly to e:
(3)$$ \begin{align} e - s_n = \frac{1}{(n+1)!} + \frac{1}{(n+2)!} + \frac{1}{(n+3)!} + \cdots < \frac{1}{(n+1)!}\bigg(1 + \frac{1}{n+1} + \frac{1}{(n+1)^2} + \cdots \bigg) = \frac{1}{n! n} \end{align} $$
Hence
(4)$$ \begin{align} 0 < e - s_n < \frac{1}{n! n} \end{align} $$
//e// is irrational. Suppose for the sake of contradiction that //e// = //p///q, where p and q are positive integers. Setting n = q in the above inequality, we we get that
(5)$$ \begin{align} 0 < q! (e - s_n) < \frac{1}{q} \end{align} $$
However
(6)$$ \begin{align} q! e &= (q-1)!p \\
q!s_q &= q!\bigg(1 + 1 + \frac{1}{2!} + \cdots + \frac{1}{q!}\bigg) \end{align} $$
which upon inspection are both integers. Thus we have shown the existence of an integer between 0 and 1, a contradiction. □
Next we prove this identity:
(7)$$ \begin{align} \lim_{n \rightarrow \infty} \bigg(1 + \frac{1}{n}\bigg)^n = e \end{align} $$
Let sn be defined as above and
(8)$$ \begin{align} t_n = \bigg(1 + \frac{1}{n}\bigg)^n \end{align} $$
Expand tn using the binomial theorem:
(9)$$ \begin{align} t_n &= 1 + 1 + \sum_{i=2}^n { n \choose i } \frac{1}{n^i} \\
&= 1 + 1 + \frac{n!}{2!(n-2)!}\frac{1}{n^2} + \frac{n!}{3!(n-3)!}\frac{1}{n^3} + \cdots + \frac{1}{n^n} \\
&= 1 + 1 + \frac{n (n-1)}{2! n^2} + \frac{n (n-1) (n-2)}{3! n^3} + \cdots + \frac{(n-1)!}{(n-1)!}\frac{1}{n^n} \\
&= 1 + 1 + \frac{1}{2!}\bigg(1 - \frac{1}{n}\bigg) + \frac{1}{3!}\bigg(1 - \frac{1}{n}\bigg)\bigg(1 - \frac{2}{n}\bigg) + \cdots + \frac{1}{n!}\bigg(1 - \frac{1}{n}\bigg)\bigg(1 - \frac{2}{n}\bigg) \cdots \bigg(1 - \frac{n - 1}{n}\bigg) \end{align} $$
This shows that tn ≤ sn and
(10)$$ \begin{align} \lim \sup_{n \rightarrow \infty} t_n \leq e \;\;\;\;\;\;\;\; (*) \end{align} $$
When m ≤ n we have
(11)$$ \begin{align} t_n \geq 1 + 1 + \frac{1}{2!}\bigg(1 - \frac{1}{n}\bigg) + \frac{1}{3!}\bigg(1 - \frac{1}{n}\bigg)\bigg(1 - \frac{2}{n}\bigg) + \cdots + \frac{1}{m!}\bigg(1 - \frac{1}{n}\bigg)\bigg(1 - \frac{2}{n}\bigg) \cdots \bigg(1 - \frac{m - 1}{n}\bigg) \end{align} $$
Fixing m and letting n → ∞ it follows that
(12)$$ \begin{align} \lim \inf_{n \rightarrow \infty} t_n \geq 1 + 1 + \frac{1}{2!} + \cdots + \frac{1}{m!} \end{align} $$
Hence
(13)$$ \begin{align} s_m \leq \lim \inf_{n \rightarrow \infty} t_n \end{align} $$
Letting m → ∞ it follows that
(14)$$ \begin{align} e \leq \lim \inf_{n \rightarrow \infty} t_n \end{align} $$
which together with (*) completes the proof. □