Proof of the irrationality of e, after Fourier

Proof

We argue by contradiction. We build an integer that lies strictly between $0$ and $1$.

1. By the series for $e$, $e = \sum_{n=0}^{\infty} \frac{1}{n!}$, where $e$ is $\exp(1)$.
2. Assume that $e$ is not irrational. Then $e = a/b$ for an integer $a$ and an integer $b \ge 1$.
3. Let $x = b! \left( e - \sum_{n=0}^{b} \frac{1}{n!} \right)$.
4. $x$ is an integer. First, $b! \, e = b! \, a / b = a \cdot (b-1)!$, an integer. Second, for $0 \le n \le b$, the definition of the factorial gives $\frac{b!}{n!} = (n+1)(n+2) \cdots b$, an integer.
5. For $N > b$, $\sum_{n=0}^{N} \frac{1}{n!} - \sum_{n=0}^{b} \frac{1}{n!} = \sum_{n=b+1}^{N} \frac{1}{n!}$. As $N \to \infty$, the left side tends to $e - \sum_{n=0}^{b} \frac{1}{n!}$, by the limit of a sum.
6. Multiply by $b!$ and use the algebra of limits. Then step 5 gives $x = \sum_{k=1}^{\infty} t_k$, where $t_k = \frac{b!}{(b+k)!}$.
7. All $t_k$ are positive. By the result on series of nonnegative terms, $x$ is the supremum of the partial sums, so $x \ge t_1 > 0$.
8. Let $q = \frac{1}{b+1}$, so $0 < q \le \frac12$. For $k \ge 1$, $t_k = \frac{1}{(b+1)(b+2) \cdots (b+k)} \le q^k$, since each factor is at least $b + 1$.
9. For $k \ge 2$, the factor $b + 2$ exceeds $b + 1$. So $t_2 < q^2$.
10. By the geometric series with first term $q$ and ratio $q$, $\sum_{k=1}^{\infty} q^k = \frac{q}{1 - q} = \frac{1}{b}$.
11. Let $d_k = q^k - t_k \ge 0$. The series $\sum d_k$ converges, with sum $\frac{1}{b} - x$, by step 10 and the limit of a sum.
12. By the result on series of nonnegative terms, $\sum d_k \ge d_1 + d_2 \ge d_2 > 0$, using step 9. So $x < \frac{1}{b} \le 1$.
13. By steps 7 and 12, $x$ is an integer with $0 < x < 1$. No such integer exists. So $e$ is irrational.