I need to calculate the limit of the following sequence:

$$\lim _ {n \to \infty} \sqrt[n]{\sin(n)}$$

where the $n$-th root of a negative number is defined as the principal complex root.

I suspect the answer to be $1$, but I do not know how to prove it.

  • $\begingroup$ That may be a hard problem to solve, depending on the fact that $\pi$ has a finite irrationality measure. However, if you want us to help you, you must tell us how $\sqrt[n]{x}$ is defined when $x<0$. $\endgroup$ – Jack D'Aurizio Nov 14 '15 at 21:23
  • $\begingroup$ $\sin(4)<0$, then $\sqrt[4]{\sin(4)}$ is not defined (in $\mathbb R$). Do you consider this sequence in $\mathbb C$ ? $\endgroup$ – Surb Nov 14 '15 at 21:23
  • $\begingroup$ @Surb, yes, this is a sequence in $\mathbb{C}$ $\endgroup$ – Rafał Cieślak Nov 14 '15 at 21:25
  • 1
    $\begingroup$ And $\sqrt[n]{-1}$ is... ? $\endgroup$ – Jack D'Aurizio Nov 14 '15 at 21:37
  • $\begingroup$ @JackD'Aurizio A complex number with modulus $1$, and argument $\pi / n $. I suspect this limit can make little to no sense with other reasonable $\sqrt[n]{x}$ definitions. $\endgroup$ – Rafał Cieślak Nov 14 '15 at 21:45

The problem boils down to proving that $\sin(n)$ cannot be too close to zero for small values of $n$.

We know that $\pi$ is a trascendental number with a finite irrationality measure. In particular, the inequality $$ \left| \pi-\frac{p}{q}\right| \leq \frac{1}{q^{10}} $$ may hold only for a finite number of rational numbers $\frac{p}{q}$, hence (since $\left|\sin x\right|\geq K\left|x-k\pi\right|$ when $x$ is close to $k\pi$, thanks to Adayah) in the general case $\left|\sin(n)\right|$ is greater than $\frac{C}{n^9}$ for some constant $C$. That is enough to ensure that the wanted limit is $1$ by squeezing.

  • $\begingroup$ Great answer! However, I think rather than Lipschitz continuity one needs quite an opposite property here: $|\sin x| \geqslant K |x - k \pi|$ whenever $x$ is close to $k \pi$. That holds because $(\sin x)' = \cos x$ which is $1$ or $-1$ (thus non-zero) for $x \approx k \pi$. $\endgroup$ – Adayah Nov 14 '15 at 22:15
  • $\begingroup$ @Adayah: you are right, I am including your observation in my answer. $\endgroup$ – Jack D'Aurizio Nov 14 '15 at 22:17
  • $\begingroup$ Also I'm sorry, but as far as I understand, the second part makes little or no sense. You can't pull $\text{Im}$ to the outside like that because $z^n$ may not be real. Besides, from the radius of convergence being $1$ one may conclude that $\displaystyle \limsup_{n \to \infty} \sqrt[n]{|\sin n|} = 1$, but $\liminf$ may still be lower. $\endgroup$ – Adayah Nov 14 '15 at 22:49
  • $\begingroup$ @Adayah: you are right again, simply amended. $\endgroup$ – Jack D'Aurizio Nov 15 '15 at 1:19

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