# For which $r$ does the series $\sum_{n=1}^{\infty}(2+\sin(\frac{n\pi}{3})) r^n$ converge?

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?