e is irrational

Prove that e is an irrational number.

Recall that $\,\mathrm{e}=\displaystyle\sum_{n=0}^\infty\frac{1}{n!},\,\,$ and assume $\,\mathrm{e}\,$ is rational, then

$$\sum\limits_{k=0}^\infty \frac{1}{k!} = \frac{a}{b},\quad \text{for some positive integers}\,\,\, a,b.$$

so

$$b\sum\limits_{k=0}^\infty \frac{1}{k!} =a$$

or

$$ b\left(1+1+\frac{1}{2} + \frac{1}{6} +\cdots \right)= a. $$

Where can I go from here?

Solution 1:

Hints.

We first show that $2<\mathrm{e}<3$ (see below), and hence $\mathrm{e}$ is not an integer.

Next, following up OP's thought, assuming $\mathrm{e}=a/b$, we multiply by $b!$ and we obtain $$ \sum_{k=0}^\infty \frac{b!}{k!}=a\cdot (b-1)! \tag{1} $$ The right hand side of $(1)$ is an integer.

The left hand side of $(1)$ is of the form $$ \sum_{k=0}^b \frac{b!}{k!}+\sum_{k=b+1}^\infty \frac{b!}{k!}= p+r. $$ Note that $p=\sum_{k=0}^b \frac{b!}{k!}$ is an integer, while $$ 0<r=\sum_{k=b+1}^\infty \frac{b!}{k!}=\frac{1}{b+1}+\frac{1}{(b+1)(b+2)}+\cdots<\sum_{k=1}^\infty \frac{1}{(b+1)^k}=\frac{1}{b}<1. $$

Note. The fact that $\mathrm{e}\in (2,3)$ can be derived from the inequalities $$ \left(1+\frac{1}{n}\right)^{\!n}<\mathrm{e}<\left(1+\frac{1}{n}\right)^{\!n+1}, $$ for $n=1$ for the left inequality and $n=5$ for the right inequality.

Solution 2:

I like the following mild variant, which is less popular. Equivalently, we prove that $e^{-1}$ is irrational. Suppose to the contrary that $e^{-1}=\frac{m}{n}$ where $m$ and $n$ are integers with $n\gt 0$. We have $$e^{-1}=\sum_{k=0}^{n}\frac{(-1)^k}{k!}+\sum_{k=n+1}^\infty \frac{(-1)^k}{k!}.$$ Multiply through by $n!$. We get that $$n!\sum_{k=n+1}^\infty \frac{(-1)^k}{k!}$$ must be an integer.

However, by the reasoning that leads to the Alternating Series Test, we have $$0\lt \left|n!\sum_{n+1}^\infty \frac{(-1)^k}{k!}\right|\lt \frac{n!}{(n+1)!}\lt 1.$$

The (small) advantage is that the estimation of the tail is easier than when we use the series expansion of $e$.