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

limits

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.


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

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