Is $\exists p\in\mathbb{N}:\frac{1}{\left| n^p \sin(n/2) \right|}$ is bounded for $n \in \mathbb{N}$?

Question

It's clear that $\frac{1}{\left| n^p \sin(n/2) \right|}$ is not bounded where $n \in \mathbb{R}$ because $\frac{1}{\left| n^p \sin(n/2) \right|} \to \infty$ as $n \to 2k\pi\ (k \in \mathbb{N}).$ However, how does the sequence behave for $n \in \mathbb{N}$? Is the sequence bounded for some $p \in \mathbb{N}$?

Answer 1

Alekseyev proved that $\frac1{n^u |\sin(n)|^v}$ converges to zero if $\mu(\pi) < 1+u/v$, and diverges if $\mu(\pi) < 1+u/v$, where $\mu(\pi)$ is the irrationality measure of $\pi$. In my case, $u = p$ and $v = 1$. It's proved by Salikhov that $\mu(\pi) < 7.6063$. Therefore, if $p \geq 9 > 8.6063$, $$\frac1{n^p |\sin(n/2)|} = \frac{2^p}{(n/2)^p |\sin(n/2)|} \to 0$$ as $n \to \infty$.