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Does there exist any $p >0$ such that \begin{equation*} \frac{1}{n^p \sin(n)} \to 0 \;,n\to+\infty \;? \end{equation*} If there is one, what's the infimum of those $p$? Is it also a minimum?

I started wondering about it since the set of limit points of $\big(\sin(n)\big)_{n \in \mathbb{N}}$ is the whole interval $[-1,1]$, so it seems interesting to quantify how fast this sequence clusters around $0$.

Bob
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    I don't remember the specifics, but I think I once saw a similar question which in the end referred to the irrationality measure of $\pi$. See e.g. here https://en.m.wikipedia.org/wiki/Liouville_number#Irrationality_measure and here https://mathoverflow.net/questions/53724/are-some-numbers-more-irrational-than-others – PhoemueX Feb 18 '22 at 16:46
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    Not an answer, but a thought: Consider the numbers $n\in \mathbb N$ that approximate $10^k\pi$, i.e. $3,31,314,31415,314149 \dots$. As such numbers grow, $\frac{1}{\sin{(n)}}$ will become arbitrarily large. Can $n^p$ outrace $\sin{(n)}$? Interesting question! – Keith Backman Feb 18 '22 at 16:55
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    This is closely related to currently open problems. For instance, it's unknown whether $\frac{1}{n^3 \sin(n)}$ converges. See here. – HallaSurvivor Feb 18 '22 at 21:37
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    It is quite easy to show that if $\mu(\pi) <p+1$ (the irrationality measure of $\pi$) then $\frac{1}{n^p \sin(n)} \to 0, n\to+\infty $, while if $\mu(\pi) >p+1$ then $\frac{1}{n^p \sin(n)} $ diverges and in particular since it is known that $\mu(\pi) \le 7.606..$ the smallest integer $p$ which today is known to work is $7$ – Conrad Feb 18 '22 at 22:01

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