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I want to use Jordan Inequality, i.e. $$\sin x\le \frac{2x}{\pi }\text,$$ but I don't know how.

Mohsen Shahriari
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Hhhi
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    The Jordan inequality is $\sin x\ge 2x/\pi$, and it only holds for $x\in[0,\pi/2]$. But $|\sin x| \le |x|$ holds for all $x$. – Milten Oct 18 '20 at 08:41

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$(\sin n)^p$ is dense in $[-1,1]$ so the general term of this series does not go to $0$ and the series diverge.

Edit: I notice that the link have the restriction $p$ integer, so this is only a partial result. The proof need to be examined if it can be extended to real values.

zwim
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EDIT: I realized that the sum is ill-defined for most real $p$. So you can apply the below to $\sum_{n=1}^\infty \left|\sin n\right|^{-p}$.

If $p\ge 0$, it trivially diverges. So I will try to show that $\{\sin^{q} n, n\in\mathbb N\}$ is dense in $[0,1]$ for $q>0$ (corresponding to $p<0$), and then zwim's answer takes us home.

I assume we know that $\{\sin n\}$ is dense in $[0,1]$. Let $x\in[0,1]$ and $\varepsilon>0$. Let $f(x)=x^\frac1q$. Then there is an $n_0\in \mathbb N$ such that $$ \sin n_0 \in f((x-\varepsilon, x+\varepsilon)\cap [0,1]) $$ because $\{\sin n\}$ is dense in $[0,1]$. But that shows that $$ \sin^q n_0\in (x-\varepsilon, x+\varepsilon)\cap [0,1] $$ so $\{\sin^q n\}$ is dense in $[0,1]$.

This seems too simple compared to zwim's link. Am I missing something here?

Milten
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$\sin n$ can be as small as you want, take $n=\lfloor k\pi\rfloor$ with $\{k\pi\}<\epsilon$. No positive value of $p$ will do.