Here is very simple "proof" of irrationality of $e$. I feel that I really miss something here, but can not find what.
Lets assume that there are finite integers $a$ and $b$ $$\frac{a}{b}=e=\sum_{i=1}^\infty\frac{1}{i!}$$
Lets say there are integers $a_{n}$ and $b_{n}$: $$\frac{a_n}{b_n}=\sum_{i=1}^n\frac{1}{i!}$$ So $$\lim_{n\to \infty}\frac{a_n}{b_n}=\frac{a}{b}$$ Now $$\frac{a_{n+1}}{b_{n+1}}>\frac{a_{n}}{b_{n}}$$ and $$a_{n+1}>\frac{a_{n}b_{n+1}}{b_{n}}$$ But $$\frac{b_{n+1}}{b_n} \ge n \implies a_{n+1}\ge a_{n}n$$ And as $a_n$ is integer, the $\lim_{n\to\infty}a_n=\infty$. So $a_n$ can not be a finite integer.
The main flaw in your argument is that $\lim_{n \to\infty}\frac{a_n}{b_n}=\frac{a}{b}$ does not imply $a_n \to a$ and $b_n \to b$ (even if all fractions are in reduced form). A simple example is $$ \lim_{n \to\infty}\frac{1 + n}{1 + 2n} = \frac 12 \, . $$
You also stated without proof that $\frac{b_{n+1}}{b_n} \ge n$ (which may be true or not).