Does $\sum_{n=1}^\infty |\sin n|^{cn}$ converge for some $c > 0$? If so, what $c$? And for the $c$ that give divergence, what about the terms of the series? Do they converge? I saw a question where it was $|\sin n|^n$ instead of $|\sin n|^{cn}$, and it was shown that the sequence diverges (hence the series diverges), hence the question. So let's assume $c > 1$.


There is no $c>0$ such that $$\sum_{n\geq 1}\left|\sin n\right|^{cn} \tag{0}$$ converges. Since $\pi$ is an irrational number, there exist an infinite number of rational numbers $\frac{p}{q}$ such that: $$ \left| \frac{\pi}{2}-\frac{p}{q}\right|\leq\frac{1}{q^2},\tag{1}$$ and $q$ is odd, so, since the sine function is a Lipschitz function, there exists an absolute constant $K>0$ such that: $$ \left|\sin m\right|\geq 1-\frac{K}{m} \tag{2}$$ holds for an infinite number of integer numbers $m\geq 1$. Proof: let we just take $m=p$.

From $\left|p-q\frac{\pi}{2}\right|\leq\frac{1}{q}$ it follows that $\left|\sin(p)-\sin\left(q\frac{\pi}{2}\right)\right|\leq\frac{1}{q}$, so $(2)$ holds for any $K>\frac{2}{\pi}$.

Since $$ a_m=\left(1-\frac{K}{m}\right)^{cm} \tag{3}$$ is an increasing sequence converging towards $e^{-Kc}$, the general term of $(0)$ is not $o(1)$, so the series is not convergent.

  • $\begingroup$ I get the first assertion, but I don't see how the second assertion follows, namely $|\sin m| \geq 1 - K/m$.. Maybe you meant $\pi/2$ instead of $\pi$ in the first assertion? If you don't have time to elaborate more I'll accept if this gets enough upvotes and/or someone can leave a comment or answer that elaborates more on your proof outline. $\endgroup$ Feb 9 '15 at 20:23
  • $\begingroup$ @user2566092: proof updated. $\endgroup$ Feb 9 '15 at 20:31
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    $\begingroup$ Thanks, I had a feeling it needed to be something more like $\pi / 2$ with odd $q$, and that assertion must certainly be true. Also the fact that $|\sin (x) - \sin(y)| \leq |x - y|$, so that you can choose $K > 2/\pi$. $\endgroup$ Feb 9 '15 at 21:15

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