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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. – Amr Dec 5 '12 at 19:24
Do you perhaps want to take the absolute value of the sine before you take the root? – Ben Millwood Dec 5 '12 at 19:40
This seems to be a problem about digits of $\pi$... – Karolis Juodelė Dec 5 '12 at 19:46
Sorry about that. You are right is with absolute value. – George Dec 5 '12 at 20:02… – Andrew MacFie Dec 5 '12 at 20:16
up vote 3 down vote accepted

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 $n$

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

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