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

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 $0$.
Thank you. I expected $(2+\sqrt{3})^n$ to approach an integer quickly, but how did you arrive at $(2-\sqrt{3})^n$ as the 'complement'? I don't think I'm expected to know any number theory to arrive at the answer. – Dever Jan 9 '12 at 6:07
Whenever one sees $a+b\sqrt{d}$, its partner $a-b\sqrt{d}$ is lurking in the background. Note that here $2-\sqrt{3}$ is just the reciprocal of $2+\sqrt{3}$. Hard to answer your question, have known this "trick" (but not in this specific context) forever. – André Nicolas Jan 9 '12 at 6:14
@Dever: One way to "see" the use of $(2 - \sqrt3 )^n$ as the complement is to write out integer approximants to the sequence $(2 + \sqrt 3)^n$: 4, 14, 52, 194, 724, and look for a pattern. Here these numbers satisfy the recurrence $A_n = 4 A_{n-1} - A_{n-2}$. Solving this recurrence gives Nicolas's formula for $A_n$. – Scaramouche Jan 9 '12 at 6:50
@Scaramouche: A useful thing to point out! And proving that $(2+\sqrt{3})^n$ and $(2-\sqrt{3})^n$ satisfy the recurrence, and therefore their sum does, is easy. Since $A_0$ and $A_1$ are integers, we conclude that $A_n$ is an integer for all $n$, giving a proof of the fact we need without using the Binomial Theorem. There are several others. – André Nicolas Jan 9 '12 at 7:18