complex numbers - How can I compute the limit of this sequence: $sqrt[n]{sin n}$?

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.

Answer

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.