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:


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!

  • $\begingroup$ It looks divergent in general, since $sin^{2n}(\alpha)$ is never negative and doesn't $\to 0$. $\endgroup$ Sep 23 '19 at 16:11
  • 2
    $\begingroup$ 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. $\endgroup$ Sep 23 '19 at 16:18
  • $\begingroup$ @ PierreCarre I meant $sin^2(\alpha)$ does not converge $\to 0$. $\endgroup$ Sep 23 '19 at 16:23
  • 1
    $\begingroup$ @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. $\endgroup$ 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 $\beta$.
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.