Adjacent sequences and the irrationality of e

1. $a_{n+1}-a_n=\frac1{(n+1)!}>0$. For $b$:
$$b_{n+1}-b_n=\frac1{(n+1)!}+\frac1{(n+1)(n+1)!}-\frac1{n\,n!}=\frac1{(n+1)!}\left(\frac{n+2}{n+1}-\frac{n+1}n\right)=-\frac1{n(n+1)(n+1)!}<0.$$

2. $b_n-a_n=\frac1{n\,n!}\to0$. The sequence $(a_n)$ is increasing and bounded above by $b_1$, so it converges, say to $\ell$; $(b_n)$ is decreasing and bounded below by $a_1$, so it converges to the same $\ell$, since the difference tends to $0$. This common limit is $e=\sum\frac1{k!}$.

3. Strict monotonicity gives $a_q<e<b_q$ for every $q\ge1$. Suppose $e=\frac pq$ with integers $p,q\ge1$. Multiplying $a_q<\frac pq<a_q+\frac1{q\cdot q!}$ by $q!$:
$$q!\,a_q<p\,(q-1)!<q!\,a_q+\frac1q.$$
Here $q!\,a_q=\sum_{k=0}^q\frac{q!}{k!}$ and $p\,(q-1)!$ are integers, and the integer $p(q-1)!$ lies strictly between the integer $q!\,a_q$ and that integer plus $\frac1q\le1$. Impossible.

Conclusion. $e$ is irrational.

Remark. A famous argument, due to Fourier. It uses nothing but adjacent sequences.