Let $a_1,a_2,a_3,\ldots$ be integers such that $1\leq a_n \leq n-1$ for $n=2,3,4,\ldots\,$ Prove that $$\sum_{n=1}^{\infty}\frac{a_n}{n!}=\frac{p}{q},\, p,q \in \mathbb{Z}, q \neq 0 \iff \exists N\in \mathbb{ N}: (n\geq N \implies a_n = n-1)$$ I have already proven $(\Leftarrow),$ but I'm having some doubts with my proof of $(\Rightarrow).$ The argument is similar to the proof that $e$ is not rational.
Proof: I'm assuming there is no loss of generality if I take $q>0.$
If $\sum \frac{a_n}{n!}= \frac{p}{q},$ then $q! \sum \frac{a_n}{n!} = p (q-1)! \in\mathbb{Z}.$
Now, $$q! \sum \frac{a_n}{n!} = q! \sum_{n=1}^q \frac{a_n}{n!} + q! \sum_{n=q+1}^{\infty} \frac{a_n}{n!} = p(q-1)!$$
Since $q!\geq n!$ for $n =1,2,\ldots, q$, and $a_i \in \mathbb{Z}$, we have that $\,q! \sum_{n=1}^q \frac{a_n}{n!} \in \mathbb{Z}$, which implies that $q!\sum_{n=q+1}^{\infty} \frac{a_n}{n!} \in \mathbb{Z}$ also.
Then, since $1\leq a_n \leq n-1$, and $q+1 \geq 2,$
$$\sum_{n=q+1}^{\infty} \frac{1}{n!} \leq \sum_{n=q+1}^{\infty} \frac{a_n}{n!}\leq \sum_{n=q+1}^{\infty}\frac{n-1}{n!}= \frac{1}{q!},$$ so $$q!\sum_{n=q+1}^{\infty} \frac{1}{n!} \leq q!\sum_{n=q+1}^{\infty} \frac{a_n}{n!}\leq q!\sum_{n=q+1}^{\infty}\frac{n-1}{n!}= 1.$$
Since $q >0, \,q!\sum_{n=q+1}^{\infty} \frac{1}{n!} > 0$ and also $\,q!\sum_{n=q+1}^{\infty} \frac{a_n}{n!} \in \mathbb{Z},$ so we are forced to have
$$q!\sum_{n=q+1}^{\infty} \frac{a_n}{n!} = 1 \text{ hence } \, \sum_{n=q+1}^{\infty} \frac{a_n}{n!} = \frac{1}{q!} = \sum_{n=q+1}^{\infty}\frac{n-1}{n!},$$
therefore $a_n = n-1$ for all $n \geq q+1.$
Is this correct?
If the 2 series in the last line converge to the same limit, can I conclude that they are equal term by term? I know in general that this is not true.
Thanks for your help!