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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!

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  • $\begingroup$ It looks divergent in general, since $sin^{2n}(\alpha)$ is never negative and doesn't $\to 0$. $\endgroup$ – herb steinberg Sep 23 '19 at 16:11
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    $\begingroup$ 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 $\alpha$.. $\endgroup$ – Shashi Sep 23 '19 at 16:12
  • $\begingroup$ @herbsteinberg If $\alpha \ne \frac{\pi}{2}+ k \pi $, the term $\sin^{2n} \alpha$ does converge to zero. $\endgroup$ – PierreCarre Sep 23 '19 at 16:18
  • $\begingroup$ @ PierreCarre I meant $sin^2(\alpha)$ does not converge $\to 0$. $\endgroup$ – herb steinberg Sep 23 '19 at 16:23
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    $\begingroup$ @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. $\endgroup$ – PierreCarre Sep 23 '19 at 16:26
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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 $\beta$.
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  • $\begingroup$ 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!! $\endgroup$ – Gibbs Sep 23 '19 at 16:47

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