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I heard this question from a professor a couple years ago. I still think about it...

Does the sequence $(a_n)_{n\in \mathbb N}$ with $$a_n=\sqrt[n]{|\sin(n)|}$$ converges ( to $1$ ) ?

I believe this question is related to rational approximation.

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    The sequence is not well defined.2012-12-05
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    Do you perhaps want to take the absolute value of the sine before you take the root?2012-12-05
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    This seems to be a problem about digits of $\pi$...2012-12-05
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    Sorry about that. You are right is with absolute value.2012-12-05
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    http://wolframalpha.com/input/?i=limit%20of%20%7Csin(n)%7C%5E(1%2Fn)%20as%20n%20goes%20to%20infinity&dataset=2012-12-05
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    @AndrewMacFie, does wolfram know n is natural? I don't think it does. In that case, the result is clearly wrong.2012-12-05
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    Well, it's obvious that sine is frequently bounded away from zero, therefore 1 is a cluster point. Now, the only way the sequence might not converge would be if it gets very close (meaning so close that not even $n$th root can make it big again) to zero infinitely often. So it's enough to analyze the way $n$ approximates $m \pi$, $m \in \mathbb Z$. I think this approximation is of polynomial order (meaning it's not better than $c/n^k$ for some $k \in \mathbb Z$) and therefore the $n$th root will win the race, eventually.2012-12-05
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    It is sure that the sequence is well-defined, since $\sin n\pi=0 \iff n=0$. For the second question, it seems to be more problematic.2012-12-05

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As has been suggested in the comments, this has to do with rational approximations of $\pi$, whose accuracy is quantified by the irrationality measure of $\pi$. The sequence converges because the irrationality measure of $\pi$, though unknown, is known to be finite.

For integers $p$ and $q$ with $q$ sufficiently large, we have

$$ |p-q\pi|\gt\frac1{q^{\mu-1}}\;, $$

where $\mu$ is any real number greater than the irrationality measure of $\pi$. Now if $q\pi$ is the integer multiple of $\pi$ closest to $n$, then $|\sin(n-q\pi)|\ge\frac2\pi|n-q\pi|$, so for sufficiently large $n$

$$ \sqrt[n]{|\sin(n-q\pi)|}\ge\sqrt[n]{\frac2\pi|n-q\pi|}\gt\sqrt[n]{\frac2\pi\frac1{q^{\mu-1}}}\ge\sqrt[n]{\frac2\pi\frac1{\left(n\pi+\frac\pi2\right)^{\mu-1}}}=\sqrt[n]{\frac2\pi}\exp\left(-\frac{\mu-1}n\log\left(n\pi+\frac\pi2\right)\right)\longrightarrow_{n\to\infty}1\;. $$