9
$\begingroup$

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.

  • 0
    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

1 Answers 1

3

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\;. $