# For $x$ being irrational, why is $\lim_{m\to\infty}\lim_{n\to\infty}[\cos(n!\pi x)]^{2m}=0$?

Let $x \in \mathbb{R} \setminus \mathbb{Q}$. What is the value of $$\lim_{m \to \infty} \lim_{n \to \infty} \left[ \cos(n!\pi x) \right]^{2m}, \qquad (m,n \in \mathbb{N})$$

The answer given is $0$. I don't understand why it is so.

• Related post: Double limit of $\cos^{2n}(m! \pi x)$ at rationals and irrationals. Found using Approach0. Other questions linked there might be of interest, too. Dec 26 '16 at 14:23
• @koolman, something is not right about your question: are you sure that it's $\lim_{m \to \infty} \lim_{n \to \infty}$ and not $\lim_{n \to \infty} \lim_{m \to \infty}$? Equivalently, are you sure that it's $[ \cos(n!\pi x) ]^{2m}$ and not $[ \cos(m!\pi x) ]^{2n}$? I bet you've inadvertently interchanged $m$ and $n$, because your question is not quite right. Dec 26 '16 at 16:46
• @AlexM. This is the original question imgur.com/a/3XjRM Dec 26 '16 at 16:53
• @AlexM. And how does that matter Dec 26 '16 at 16:53

[This answer has been rewritten thanks to the helpful criticism in Alex's comment.]

The statement is wrong. One has to take into consideration first the existence of the limit $$\lim_{n\to\infty}\cos(n!\pi x)^{2m},$$ which is a highly non-trivial problem.

When $x=\dfrac{1}{\pi}$, $\cos(n!\pi x)=\cos(n!).$ But it seems that the existence of limit (which could be $1$) $$\lim_{n\to\infty}\cos(n!)\qquad\text{(which could be 1)},$$ is an open problem (edited due to Fimpellizieri's comment) thanks to the answer of this question: Is there a limit of cos (n!)?.

[Added: Interestingly, there is related question in MO: On the behaviour of $\sin(n!\pi x)$ when $x$ is irrational.]

If you were to ask the double limit related to the Dirichlet function on the other hand: $$\lim_{m\to\infty}\lim_{n\to\infty}\big[\cos(m!\pi x)\big]^{2n}.$$ you could read the accepted answer to this question:

Note carefully (again, thanks to Alex's comment) that the order of taking the double limit is different from the one in your question: $$\lim_{m\to\infty}\lim_{n\to\infty}\big[\cos(m!\pi x)\big]^{2n}.$$

• How can we say this ?? Dec 25 '16 at 16:43
• No , I don't know that Dec 25 '16 at 16:50
• ... Then you might either ask a follow-up question in a new post or look it up in your calculus textbook.
– user9464
Dec 25 '16 at 16:52
• @AlexM.: I understand that you could have given your own answer without writing any comment to my previously stupid wrong hint. However you provided me a hint to learn something I didn't fully understand since I naively thought OP is asking about the Dirichlet function. Thank you very much and I apologize for my unreasonable complaint.
– user9464
Dec 26 '16 at 0:53
• The linked answer does not state that $\cos(n!)\to1$, but rather that there is an (unlikely) scenario, which we do not yet know if it is or not true, in which that limit holds. Dec 26 '16 at 0:59

Let $e=\sum_{j=0}^{\infty}j!^{-1},$ which is irrational, because $n!e\not \in \mathbb Z$ for any $n\in \mathbb N.$

For $n\in \mathbb N$ we have $$n!e=A_n+\sum_{j=n+1}^{\infty}n!/j!=A_n+B_n,$$ where $A_n\in \mathbb N$ and $B_n\in (0,1)$.By induction on $n,$ if $n$ is even then $A_n$ is odd, and if $n$ is odd then $A_n$ is even.

So for infinitely many $n$ we have $\cos (\pi n!e)=\cos \pi B_n$ and for infinitely many $n$ we have $\cos (\pi n!e)=-\cos \pi B_n.$

Therefore, since $\lim_{n\to \infty }B_n=0,$ the sequence $(\cos n!\pi e)_{n\in \mathbb N}$ has a $\lim \sup$ of $+1$ and a $\lim \inf$ of $-1.$ Therefore $\lim_{n\to \infty}\cos(n!\pi e)$ does not exist.

• @koolman, if I had been you, then this is the answer that I would have accepted. Dec 26 '16 at 16:39