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

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!

• It looks divergent in general, since $sin^{2n}(\alpha)$ is never negative and doesn't $\to 0$. – herb steinberg Sep 23 '19 at 16:11
• Maybe set $x=2\sin^2(\alpha)$ and ask yourself for which values of $x\geq 0$ and $\beta$ the series $\sum \frac{x^n} {n^\beta}$ converges. After that translate back to $\alpha$.. – Shashi Sep 23 '19 at 16:12
• @herbsteinberg If $\alpha \ne \frac{\pi}{2}+ k \pi$, the term $\sin^{2n} \alpha$ does converge to zero. – PierreCarre Sep 23 '19 at 16:18
• @ PierreCarre I meant $sin^2(\alpha)$ does not converge $\to 0$. – herb steinberg Sep 23 '19 at 16:23
• @Gibbs I think Shashi's comment you surely get you to the correct answer. The upper bound you obtained was just too much. Bounding a series by a divergent series does not allow you to draw any conclusions. – PierreCarre Sep 23 '19 at 16:26

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 $$\beta$$.
• I posted a comment with my own solution, but I made a wrong assumption (with $x\leq1$ and $\beta\leq1$). Using ratio test seems to be the best way to solve the problem. Thanks!! – Gibbs Sep 23 '19 at 16:47