Irrationality of \(e\)

Below is a pretty standard proof of the irrationality of \(e\), which depends on the series representation of \(e\) from evaluating the Taylor series expansion of \(\exp\) about \(0\) at \(1\).

Theorem

\(e\) is irrational.

Lemma

Setting \(e_n = \sum_{k = 0}^n \frac{1}{k!}\) we have that

\[ e - e_n < \frac{1}{n \cdot n!}\]

for all \(n \in \mathbb{N}\).

Proof

The idea of this bound is to multiply through by \(n!\) to bound in terms of a geometric series.

\[\begin{align*} e - e_n &= \sum_{k = 0}^\infty \frac{1}{k!} - \sum_{k = 0}^n \frac{1}{k!} \\ &= \sum_{k = n + 1}^\infty \frac{1}{k!} \\ &= \frac{1}{n!} \sum_{k = n + 1}^\infty \frac{n!}{k!} \\ &= \frac{1}{n!} \left(\frac{n!}{(n + 1)!} + \frac{n!}{(n + 2)!} + \frac{n!}{(n + 3)!} + \dots \right) \\ &= \frac{1}{n!} \left(\frac{1}{n + 1} + \frac{1}{(n + 1)(n + 2)} + \frac{1}{(n + 1)(n + 2)(n + 3)} + \dots \right) \\ &< \frac{1}{n!} \left(\frac{1}{n + 1} + \frac{1}{(n + 1)^2} + \frac{1}{(n + 1)^3} + \dots \right) \\ &= \frac{1}{n!} \left(\frac{1}{n + 1} \left(\frac{1}{1 - \frac{1}{n + 1}}\right) \right) \\ &= \frac{1}{n!} \left(\frac{1}{n + 1} \left(\frac{n + 1}{n}\right) \right) \\ &= \frac{1}{n \cdot n!} \\ \end{align*}\]

We can use this bound to then construct a contradiction on \(e\) being rational.

Proof

From the previous part we know that \(e- e_n < \frac{1}{n \cdot n!}\) and hence for \(n > 1\) we have

\[ 0 < n! (e - e_n) < \frac{1}{n} < 1.\]

Now, if we assume that \(e\) is rational, that is \(e = \frac{a}{b}\) for integers \(a\) and \(b\), we can deduce that

\[\begin{align*} n! (e - e_n) &= n! \cdot \frac{a}{b} - n! \sum_{k = 0}^n \frac{1}{k!} \\ &= a \cdot \frac{n!}{b} - \sum_{k = 0}^n \frac{n!}{k!}. \\ \end{align*}\]

Now, because \(k < n\) in the right hand sum, we have that \(k! \mid n!\) and thus this term is an integer. Similarly for \(n \geq b\), the left hand term is also guaranteed to be an integer. However this contradicts the previous constraint of \(n! (e - e_n)\) being strictly between two adjacent integers.