TITLE: Closeness of $n! \ x$ to integers for irrational $x$
QUESTION [5 upvotes]: 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
$$
  \lim_{n\to\infty} | a_n - n! \ x | = 0.
$$

REPLY [4 votes]: 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
$$
  a_n = \sum_{k=0}^n \frac{n!}{k!}.
$$