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One of my teachers have given a limit to compute:

$$\displaystyle\lim_{n\to\infty}|\sin n|^{\frac{1}{n}}$$

I have proved that if the limit exits it has to be $1$. (By using the fact that $\{n\pi\}$ is dense in $[0,1]$)

But I seem to have no idea how to approach the problem from here. Any help would be appreciated.

Grobber
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    By using the fact that{$n \pi$} is dense. where it is dense? – GA316 Aug 22 '14 at 15:52
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    I have used the fact that ${n\pi}$ can be made as close to $1$ as desired. – Grobber Aug 22 '14 at 16:02
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    @Grobber I'd use the fact that $\forall n\in\mathbb{N}_0$ $0<|\sin{n}|<1$, and since the argument will make the logarithm negatively finite, it will always be $o(n)$ as $n\to+\infty$. – TheVal Aug 22 '14 at 16:08
  • An approach: is there a subsequence of ${n\pi}$ that approaches $0$ so quickly that $({n\pi})^n\to L<1$? – 2'5 9'2 Aug 22 '14 at 16:11
  • Dirty Rough answer: $$ \lim_{n\to\infty}|\sin n|^{\frac{1}{n}} \= x^{\frac1{\infty}} ,\quad x\in [0,1]\= x^0 = 1$$

    Notice how the value of $x$ just oscillates in the interval :D

     – Nick Aug 22 '14 at 16:18
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    @Nick a sequence of numbers $x_n$ that converges to $0$ quickly enough can have $\lim_{n \to \infty} x_n^{1/n} = 0$. – user2566092 Aug 22 '14 at 16:28

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Use this result: $$m>1,m<n,|n-m\pi|<\dfrac{\pi}{2}$$

then have $$|n-m\pi|>\dfrac{1}{m^{41}}$$ this reslut proof can see K.Mahler.on the approximation of $\pi$,Indag.Math.

so $$1>|\sin{n}|=|\sin{(n-m\pi)}|>\dfrac{2}{\pi}|n-m\pi|>\dfrac{2}{\pi\cdot n^{41}}$$

beep-boop
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math110
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  • That's a pretty crazy exponent of $41$. How hard is the result to prove? This is a question posed by a teacher, maybe there is an easier way to get a worse bound that still works to evaluate the limit? Or maybe the teacher didn't realize how hard the question was? – user2566092 Aug 22 '14 at 16:34
  • If $m$ and $n$ are both positive integers and $m>n$, then $m-n \geq 1$, which means $\pi m - n \geq 1 + (\pi-1)m > \pi$, then how can be $|n-m\pi| < \frac{\pi}{2}$? – Petite Etincelle Aug 22 '14 at 16:46
  • @LiuGang,sorry,it's $m<n$ – math110 Aug 22 '14 at 16:54
  • @math110 ok, so essentially you said the speed that a subsequence of $n$ approaches integer times $\pi$ can't be faster than polynomial – Petite Etincelle Aug 22 '14 at 17:14
  • @math110 I think is $42$ not $41$. – Krokop Aug 22 '14 at 17:16
  • @math110 Please edit your answer, is $42$! – Krokop Aug 23 '14 at 17:12
  • @math110 Could you please give more information regarding the article you mention? For instance, a link, or some more explicit reference. – Aloizio Macedo Sep 16 '19 at 21:39
  • @AloizioMacedo https://carma.newcastle.edu.au/resources/mahler/docs/119.pdf – Bcpicao Nov 25 '20 at 11:41