Does $\sum_{n=1}^{\infty} \sin(\pi(2+\sqrt{3})^n)$ converge? Converge absolutely? [duplicate]

Per the title, does $\displaystyle\sum_{n=1}^{\infty} \sin(\pi(2+\sqrt{3})^n)$ converge? Converge absolutely?

I'm stuck on this question, not sure how to approach it.

Thank you!

Hint: Let $$A_n=(2+\sqrt{3})^n+(2-\sqrt{3})^n.$$ Show first that $A_n$ is an integer. There are various ways of seeing this, including expanding both powers using the Binomial Theorem, and noting the cancellations. Or else, if you are familiar with recurrences, the $A_n$ satisfy a nice Fibonacci-like recurrence relation.

But $0<2-\sqrt{3}<0.27$, so $(2-\sqrt{3})^n$ goes to $0$ rapidly as $n$ gets large. Thus, for large $n$, $(2+\sqrt{3})^n$ is only a tiny bit below an integer. That gives us a good handle on the size of $\sin(\pi(2+\sqrt{3})^n)$. For $$\sin(\pi(2+\sqrt{3})^n)=\sin(\pi A_n -\pi(2-\sqrt{3})^n).$$ Now use the formula $\sin(x+y)=\sin x\cos y +\cos x\sin y$, and standard estimates for $|\sin t|$ when $t$ is close to $0$.