How to show that $e$ is irrational by studying $\sum^{n}_{k=1}\frac{1}{k!}$ and $\sum^{n}_{k=0}\frac{1}{k} + \frac{1}{n \cdot n!}$?

Question

Suppose you have $$a_n = \sum^{n}_{k=0}\frac{1}{k!}$$ and $$b_n = a_n + \frac{1}{n \cdot n!}. $$

By using the fact that $a_n <e <b_n$, which is true $\forall n \in \mathbb{N}$, conclude that $e \not \in \mathbb{Q}$.

And I am unable to do anything. I tried to get something absurd by supposing that $e = \frac{a}{b}$ with $a,b \in \mathbb{Z}$, but didn't get anywhere. A hint would be highly appreciated.

Answer 1

Assume that $e=p/q$ with $p,q\in \mathbb{N}^+$ then for $n=q$, $$(qq!)\cdot a_q <(qq!)\cdot e <(qq!)\cdot b_q\implies qq!a_q<pq(q-1)!<qq!a_q+1.$$ which means that the integer $pq(q-1)!$ is strictly between the two consecutive integers $qq!a_q$ and $qq!a_q+1$. Contradiction.

P.S. $(qq!)\cdot a_q$ is an integer because $$(qq!)\cdot a_q=q\sum^{q}_{k=0}\frac{q!}{k!}= q\sum^{q-1}_{k=0}\prod_{k+1}^q j+q$$ which is a sum of integers.

Answer 2

Hint: Let $n=b$, and clear the denominators in the inequality $a_n<e<b_n$.