Hi, I'm interested in the behaviour of the sequence $(\sin(n!\pi x))$, when $x$ is irrational, as $n$ tends to infinity.
1) Is the sequence dense in $(-1,1)$?
or
2) Is it possible that for some irrational $x$, $\sin(n!\pi x)$ tends to $0$ as $n$ tends to infinity?
Any reference would be appreciated,
Thank you
For $x=e=\sum 1/{{i}!}$, the sequence $sin(n!\pi x)$ tends to zero since the sequence of fractional parts {$n!x$} tends to zero. Hence generally answer on your question is negative.
Denote by G the set of all x for which $sin(n! \pi x)$ approaches 0 as n approaches infinity. Every real number $0 < x < 1$ can be uniquely written as $ \sum_{n \ge 2} \frac{x(n)}{n!}$ where $0< x(n) < n$. From this it can be immediately seen that $x \in G$ iff $\frac{x(n)}{n}$ approaches 0 or 1. This immediately shows that there are continuum many points of G in every interval.
It can also be shown that G is an F-sigma-delta (countable union of countable intersections of open sets) additive subgroup of reals but it's neither G-delta nor F-sigma. One may iterate this construction to obtain additive groups of reals at arbitrarily high finite levels of Borel hierarchy in the following way: Let $G_0 = G$. Let $G_{n+1}$ be the set of all x such that the fractional part of $(n!x)$ converges to a point in $G_n$. Then $G_n$'s form an increasing chain of additive subgroups of reals and their union is a Borel additive subgroup of reals which is not at any finite level of Borel hierarchy.
I wrote a note on this here.
