Is the series $\sum \sin^n(n)$ divergent?

I'm almost sure that the series $\sum \sin^n(n)$ is not convergent, but lack proof. Thank for any help.

Using a continued fraction approximation, we know that for any positive integer $N$, we can find integers $q>N$ and $p$ so that $$ \left|p-q\frac\pi2\right|<\frac1q\tag{1} $$ It is also true that no two consecutive denominators can share a common factor. Therefore, if one approximation has an even denominator, the next must be odd. So assume that $q$ is odd and $(1)$ is true. Then, since $\sin\left(q\frac\pi2\right)=(-1)^{(q-1)/2}$ and $\cos\left(q\frac\pi2\right)=0$ the Maclaurin Series yields

$$ \begin{align} (-1)^{(q-1)/2}\sin(p) &\ge1-\frac12\left(p-q\frac\pi2\right)^2\\ &\ge1-\frac{1}{2q^2}\tag{2} \end{align} $$ Thus, for continued fraction approximations $\frac{p}{q}$ to $\frac\pi2$ with odd denominators $|\sin(p)|\ge1-\frac{1}{2q^2}$. Taking the $\liminf$ as $q\to\infty$ yields $$ \begin{align} \liminf_{p\to\infty}|\sin(p)|^p &\ge\lim_{p\to\infty}\left(1-\frac{1}{2q^2}\right)^p\\ &=\lim_{q\to\infty}\left(1-\frac{1}{2q^2}\right)^{q\pi/2}\\ &=1\tag{3} \end{align} $$ Inequality $(3)$ implies that $$ \limsup_{n\to\infty}|\sin^n(n)|=1\tag{4} $$ Since the terms do not tend to $0$, $$ \sum_{n=0}^\infty\sin^n(n) $$ does not converge.

For any irrational $r$ (in particular $\pi/2$) there are infinitely many fractions $p/q$ such that $|r - p/q| < 1/q^2$. IIRC we can specify that $q$ is odd, perhaps at the cost of changing $1/q^2$ to $c/q^2$ for some constant $c$. Taking $r = \pi/2$, this says $|q \pi/2 - p| < c/q$, and then $|\sin p| > 1 - c^2/(2 q^2)$ and $|\sin p|^p > (1-c^2/(2 q^2))^p$. Now as $q \to \infty$ with $p \approx q \pi/2$, $(1 - c^2/q^2)^p \to 1$. Thus for any $\epsilon > 0$ there are infinitely many $n$ with $|\sin n|^n > 1 - \epsilon$.

Hint: If $\frac{n}{k}$ is a "good" approximation to $\frac{\pi}2$ and $k$ is odd and large, then $|\sin^n n|$ is close to $1$.