# How to prove that $\lim(\underset{k\rightarrow\infty}{\lim}(\cos(|n!\pi x|)^{2k}))=\chi_\mathbb{Q}$ [duplicate]

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How is this called? Rationals and irrationals

Please help me prove, that $$\underset{n\rightarrow\infty}{\lim}\left(\underset{k\rightarrow\infty}{\lim}(\cos(|n!\pi x|)^{2k})\right)=\begin{cases} 1 & \iff x\in\mathbb{Q}\\ 0 & \iff x\notin\mathbb{Q} \end{cases}$$

Seems very complicated, but it's on calc I. I've tried use series expansions of cos, but it don't lead to answer. Thanks in advance!

Edit

• is that $\pi$, or really $\Pi$? Commented Jan 11, 2013 at 19:10
• just Pi = 3.1415... Commented Jan 11, 2013 at 19:12
• +1 Nice question. Also, the absolute value isn't really necessary, since $\cos$ is an even function. Commented Jan 11, 2013 at 19:14
• I know that I’ve answered this or a very similar question here before, but it may take a while to find it. Commented Jan 11, 2013 at 19:16

• because $cos(m!\pi x) < 1$ when $x$ is irrational, so if you raise it to large powers, it goes to 0. Commented Jan 11, 2013 at 19:21
Hint: Show that if $x \in \mathbb{Q}$, then there exists some $N$ such that for $n > N$, $n! \pi x$ is an integer multiple of $2\pi$. Conclude that it tends to 1.
Show that if $x \not \in \mathbb{Q}$, then $\lim_{k\rightarrow \infty} [\cos (n! \pi x)]^{2k} = 0$.
• So then x = p/q. When N > 2q, $2\mid n!\pi x$. Could You advice more about the second part? Commented Jan 11, 2013 at 19:17