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We have to split this up by cases based on $r$.

1) Suppose that $0<|r|<1$. Then $|(2+\sin(\frac{2\pi}{3}))r^n| \le |3r^n|$. But $0<|r|<1$, so the series $\sum_{n=1}^{\infty}3r^n$ converges. By the direct comparison test, the series ${\sum_{n=1}^\infty(2+\sin)\frac{n\pi}{3})) r^n}$ converges absolutely.

2) Suppose that $r=1$. Then $a_n = 2+\sin(\frac{2\pi}{3})$. This sequence oscillates, so does not converge to 0. Hence the series diverges.

3) Suppose that $r=-1$. Then let $a_n = (-1)^n(2+\sin(\frac{n\pi}{3}))$. This sequence oscillates, so does not converge to 0. Hence the series diverges.

4) Suppose that $r>1$. Consider the subsequence $a_{6k-5} = (2+\frac{\sqrt{3}}{2})r^n$. For this subsequence $\lim_{k\to\infty}a_{6k-5} = +\infty$. Hence $\lim_{n\to\infty}a_n \not = 0$. So the series diverges.

5) Suppose that $r<-1$. Let $p=-r$. Then $a_n=(-1)^n(2+\sin(\frac{2\pi}{3}))p^n$. Consider the subsequence $a_{6k-5} = (-1)^n(2+\frac{\sqrt{3}}{2})p^n$. The limit of this sequence does not exist, because it oscillates to positive and negative infinity. Hence $\lim_{n\to\infty}a_n \not = 0$. Hence the series diverges.

So here is my question. Is this the best way to do this? Is there a slicker way to do it without all of the case work?

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