# Is the existence of $\lim_{n\to\infty}\cos(n!\pi x)$ for given arbitrary irrational $x$ an open problem?

Motivated by a recent MSE question about the sequence of function $\cos(n!\pi x)$, I have read related several related questions:

Is the existence of $\lim_{n\to\infty}\cos(n!\pi x)$ for given arbitrary irrational $x$ still an open problem? Are there any known related references/results for this question?

• However, $(\sin(n!\pi x))^2+(\cos(n!\pi x))^2=1$, which relates those two sequences quite closely. – Vladimir Dotsenko Dec 26 '16 at 2:50
• @T.Amdeberhan depends what $y$ is; in full generality it is not true, of course. (If $y_n=\frac{\pi}2+\pi n$, then the limit of $\cos(y_n)$ exists and is $0$, while $\sin(y_n)$ obviously has no limit.) – Vladimir Dotsenko Dec 26 '16 at 6:21
• For any $a\in[-1,1]$, there exist values of x such that $cos(n!x)\to a$. This argument is well known in ergodic theory and is essentially due to Pollington. The point is that $n!$ is a lacunary sequence (I.e. The ratio between successive terms is bounded away from 1) and in this case converges to infinity. – Anthony Quas Dec 26 '16 at 15:29
• @Jack: For the version of the problem using the sine function, we have answers by Petya and Ashutosh demonstrating a particular technique. Have you tried applying that technique to the version using the cosine? If so, then tell us what happened. If not, then it seems premature to post the question. – Ben Crowell Dec 26 '16 at 15:31
• @Jack: I don't see much connection between the linked question and mine. Your question seems to differ from that one in two ways: (1) you changed sine to cosine, and (2) you seem to be asking for an algorithm or decision procedure that determines convergence for a given $x$. What is the motivation for changing both of these things at the same time? Re #2, it's not clear to me what it would mean to have a decision procedure that took an "arbitrary irrational" as an input, since only countably many irrationals can even be described. This question seems to lack motivation and research effort. – Ben Crowell Dec 26 '16 at 19:20

Weyl proved in 1916 (see Satz 21 on Page 348 in his paper) that if a sequence of real numbers $\lambda_1<\lambda_2<\dots$ grows sufficiently rapidly (and $\lambda_n=n!$ satisfies the precise constraint there), then for almost every real number $x$ (in the sense of Lebesgue measure), the sequence of fractional parts $\{\lambda_n x\}$ is dense in $[0,1]$.
As a consequence, for almost every $x\in\mathbb{R}$, the limit of $\cos(n!\pi x)$ does not exist. This answers your first question. I don't know the answer to your second question, but I expect that the limit of $\cos(n!)$ does not exist either.