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I am having trouble calculating the following limit: $\lim_{n \to \infty} \sqrt[n]{|\sin n|}\ .$

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    @Corel: Well, I did say a *sophisticated* troll...But in any case, your updated answer has proved me wrong!2011-06-27

3 Answers 3

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Hint:

$\pi$ is not a Liouville number, so there exists $m\in\mathbb{N}$ such that for all $p,q\in\mathbb{Z}$ with $q>1$, we have $ \left| \pi - \frac{p}{q}\right| \geq \frac{1}{q^m}.$ This should allow you to keep $\sin n$ away from 0.


Edit: Full Solution:

Let $m$ be as above. So for all $p,q\in\mathbb{Z}$ with $q>1$ we have $ \left| \pi - \frac{p}{q}\right| \geq \frac{1}{q^m}.$ Now take $n\in\mathbb{N}$. Take $q_n$ so that $|q_n\pi-n|$ is minimized. Then we have $ \frac{\pi}{2} \geq \left| q_n\pi - n\right| \geq \frac{1}{q_n^{m-1}}.$ Next we note that $|\sin n| = |\sin(q_n\pi-n)|$. Now since $\sin$ is increasing on $[0,\pi/2]$ we have $|\sin(q_n\pi-n)|\geq \sin\frac{1}{q_n^{m-1}} \geq \frac{1}{2}\frac{1}{q_n^{m-1}}.$ Such an estimate holds for each $n$, with $q_n\approx \frac{n}{\pi}$. So now we have $\frac{1}{2}\frac{1}{q_n^{m-1}} \leq |\sin n| \leq 1.$ Now take $n$th roots of everything and let $n\rightarrow\infty$. The LHS goes to 1 so $\sqrt[n]{|\sin n|} \rightarrow 1.$

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    Yes, this looks good indeed.2011-06-27
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If the limit exists, then it should be equal to 1, because $|\sin n|$ is dense in $[0,1]$ and there exists a subsequence $|\sin n_k|$ converging to $1$. Then

$\sqrt[n_k]{|\sin n_k|} \to 1.$

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    The point is, does the limit exist?2011-06-27
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HINT:

There are arbitrarily large multiples of $\pi$, and there are arbitrarily large odd multiples of $\pi /2$.

Of course, this only happens if we consider real n as opposed to natural n...

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    I don't see how this answers the question at all.2011-06-27