Fractional part of the product of a factorial tending to infinity and of an irrational number Let $\alpha\in\mathbb{R}\setminus\mathbb{Q}$ be an irrational number and take the limit
$$\lim_{n\to\infty}\cos^{2n}(n!2\pi \alpha).$$
Intuitively this must be zero since $n!\alpha$ will never be an integer and therefore the value or $|\cos|$ will always be less than 1, and taking an increasingly high power of values strictly less than 1 should tend to 0. That's my intuition.
But how can I be sure that in the infinite set of values of $n!x$ there will not be a sequence of values $(n_i!x)_i$ such that their fractional values are increasingly close to $0$ that even when taking the $2n$-th powers of the cosinus, the proximity to zero will compensate the effect of the power and the result will tend to something different than $0$?
Ideally I would like to find an upper bound $\eta$ such that $|\cos(n!2\pi \alpha)|<\eta<1$ for $n$ big enough, so that $\lim_{n\to\infty}\cos^{2n}(n!2\pi \alpha)\leq \lim_{n\to\infty}\eta^{2n}$ which is necessarily $0$. But does such a bound exist?
What do we know of the fractional part of $n!\alpha$ when $\alpha$ is an arbitrary irrational and $n\to\infty$?

You can get a counterexample by taking $\alpha$ to be Euler's number $e$. We have
$$
n!2\pi e=n!2\pi\sum_{k=0}^\infty\frac{1}{k!} = 2\pi\sum_{k=0}^n \frac{n!}{k!} + 2\pi \sum_{k=n+1}^\infty \frac{n!}{k!},
$$
which differs from an integer multiple of $2\pi$ by
$$
2\pi \sum_{k=n+1}^\infty \frac{n!}{k!} = 2\pi\sum_{k\geq 1}\frac{1}{(n+1)\cdots (n+k)}< 2\pi\sum_{k\geq 1}\frac{1}{(n+1)^k}= \frac{2\pi}{n}.
$$
Since $\cos(x) \geq 1-x^2/2$, it follows that
$$
\cos(n!2\pi e)\geq 1-\frac{4\pi^2}{n^2},
$$
and therefore
$$
\begin{align*}
1\geq \cos^{2n}(n!2\pi e)&\geq \left(1-\frac{4\pi^2}{n^2}\right)^{2n}\\
&=\left(\left(1-\frac{4\pi^2}{n^2}\right)^{n^2}\right)^{2/n}.\\
\end{align*}
$$
Finally, taking the limit as $n\to\infty$, we have
$$
\lim_{n\to\infty} \left(1-\frac{4\pi^2}{n^2}\right)^{n^2} = e^{-4\pi^2},
$$
and so
$$
1 \geq \lim_{n\to\infty}\cos^{2n}(n!2\pi e)\geq  \lim_{n\to\infty} \left(e^{-4\pi^2}\right)^{2/n} = 1.
$$
