# 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 $$n\to\infty$$?

• In case what you really wanted is this one: Double limit of $\cos^{2n}(m! \pi x)$ at rationals and irrationals
– user996848
Commented Nov 24, 2021 at 15:04
• It works in their case because they take a double limit. I have only one limit. Maybe it is not possible to show that the limit is 0 with a single limit. Commented Nov 24, 2021 at 15:19
• Well, there is a big difference in the double one: the inner limit is $\lim_{n\to\infty}\cos^{2n}(m!\pi x)$; but you have $\lim_{n\to\infty}\cos^{2n}(n!\pi x )$.
– user996848
Commented Nov 24, 2021 at 16:29

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 $$1 \geq \lim_{n\to\infty}\cos^{2n}(n!2\pi e)\geq \lim_{n\to\infty} \left(e^{-4\pi^2}\right)^{2/n} = 1.$$