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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.

Martin Sleziak
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Koolman
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2 Answers2

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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.

DanielWainfleet
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[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:

Double limit of $\cos^{2n}(m! \pi x)$ at rationals and irrationals.

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

Community
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  • How can we say this ?? – Koolman Dec 25 '16 at 16:43
  • Do you know for instance why $\lim_{m\to\infty}\left(\frac{1}{2}\right)^{2m}=0$? –  Dec 25 '16 at 16:46
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    No , I don't know that – Koolman Dec 25 '16 at 16:50
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    ... Then you might either ask a follow-up question in a new post or look it up in your calculus textbook. –  Dec 25 '16 at 16:52
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    @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. –  Dec 26 '16 at 00:53
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    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. – Fimpellizzeri Dec 26 '16 at 00:59
  • @Jack: Your answer doesn't clarify why the result should be $0$, as per the OP's claim. My belief is that the OP inadvertently interchanged $m$ and $n$. The fact that he has accepted your answer means nothing, because prior to this he had accepted another one. :) – Alex M. Dec 26 '16 at 16:49
  • @AlexM.: No. As I wrote in the answer, to the best of my knowledge, the existence of the limit $\lim_{n\to\infty}\cos(n!\pi x)$ for $x$ being irrational in general is an open problem and thus it does not make sense to ask the double limit in OP without first giving the existence for all irrationals. That's it. In the second part of my answer, it is pointed out that OP might be asking the double limit regarding the Dirichlet function, which is a duplicate and I have given links to both the question and answer. –  Dec 26 '16 at 17:39
  • @Jack: True that. With so mant events happening on this page, I forgot about the second part of your post. – Alex M. Dec 26 '16 at 17:42