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!
Best 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: