e

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 tnsn and

(10)
\begin{align} \lim \sup_{n \rightarrow \infty} t_n \leq e \;\;\;\;\;\;\;\; (*) \end{align}

When mn 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. □

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