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 $n\to\infty$?

  • $\begingroup$ In case what you really wanted is this one: Double limit of $\cos^{2n}(m! \pi x)$ at rationals and irrationals $\endgroup$
    – user996848
    Commented Nov 24, 2021 at 15:04
  • $\begingroup$ 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. $\endgroup$
    – yannis
    Commented Nov 24, 2021 at 15:19
  • $\begingroup$ 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 $\lim_{n\to\infty}\cos^{2n}(n!\pi x )$. $\endgroup$
    – user996848
    Commented Nov 24, 2021 at 16:29

1 Answer 1


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


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