sequences and series - Convergence of $sum_{n=1}^inftyfrac{2^nsin^{2n}(alpha)}{n^beta}$

Determine for which values of the parameters $\alpha,\beta\in\mathbb{R}$ the following series is convergent: $$\sum_{n=1}^\infty\frac{2^n\sin^{2n}(\alpha)}{n^\beta}$$

It seems clear that if $\alpha=\pi k, k\in\mathbb{Z},$ then $\forall\beta$ the series converges as $\sin^{2n}(\alpha)=0$. Otherwise, as $0\leq\sin^{2n}\leq1$ we can fit the series in the following way:

$$\sum_{n=1}^\infty\frac{2^n\sin^{2n}(\alpha)}{n^\beta}\leq\sum_{n=1}^\infty\frac{2^n}{n^\beta}$$

But I don't know how to continue. The right term of the inequality is always divergent, so I can't apply comparison. Could you give me some hints? Thanks in advance!

Answer

Denote $\gamma = 2 \sin^2(\alpha)$. We have to study the convergence of the series $\sum u_n(\gamma, \beta)$ where $u_n(\gamma, \beta) = \frac{\gamma^n}{n^\beta}$.

Easy case... $\gamma = 0$ or $\alpha = k \pi$ with $k \in \mathbb Z$. The general term of the series is equal to zero, so the series converges.

So let's suppose that $\gamma \neq 0$ and separate the cases:

1. $\vert \sin(\alpha)\vert= 1/\sqrt{2}$, then $ \gamma = 1$ and $u_n(\gamma, \beta) = 1/n^\beta$. The series converges for $\beta >1$ and diverges otherwise.


2. $\vert \sin(\alpha)\vert \neq 1/\sqrt{2}$, then $\left\vert \frac{u_{n+1}(\gamma, \beta)}{u_n(\gamma, \beta)} \right\vert = \gamma \left(\frac{n+1}{n}\right)^\beta$. According to the ratio test, the series converges for $\gamma <1$ and diverges for $\gamma >1$ whatever the value of $\beta$.