Closeness of $n! \ x$ to integers for irrational $x$

Question

This question came up in the comments to another question.

Is there an irrational number $x$ such that, for sufficiently large $n$, the product $$ n! \ x $$ is arbitrarily close to an integer?

More formally: does there exist an irrational number $x$ and an integer sequence $(a_n)$ such that $$ \lim_{n\to\infty} | a_n - n! \ x | = 0. $$

Answer 1

Yes, the number $e=\exp(1)$ is such a number. Also, maybe more famously $\exp(-1)$.

Answer 2

I have accepted Marc van Leeuwen's answer. For those who want to see more details (as did I):

We have $$ e = \exp(1) = \sum_{k=0}^\infty \frac 1 {k!}, $$ and hence $$ n! \ e - n! \sum_{k=0}^\infty \frac 1 {k!} = 0, $$ from which we get $$ n! \ e - a_n = \sum_{k=n+1}^\infty \frac 1 {k!} $$ with the integer sequence $$ a_n = \sum_{k=0}^n \frac{n!}{k!}. $$