How to investigate the $\limsup$, the $\liminf$, the $\sup$, and especially the $\inf$ of the sequence $(\sqrt[n]{|\sin{n}|})_{n=1}^{\infty}$?

Question

How to investigate the $\limsup$, the $\liminf$, the $\sup$, and especially the $\inf$ of the sequence $(\sqrt[n]{|\sin{n}|})_{n=1}^{\infty}$?

Edit: The limit of this sequence is already investigated years ago in this post: Calculate $\lim_{n \to \infty} \sqrt[n]{|\sin n|}$. So the $\limsup$, the $\liminf$, and the $\sup$ are clearly 1. Sorry for did not search wisely.

Answer 1

Well first of all try to calculate the absolute value of |sinn|. Since the sine function is defined on (-1;1) the absolute value of |sinn|<1. Then the order of the root, as it tends to infinity the root will get smaller and smaller so basically it will tend to 0. Try to figure out yourself the interval of definition of the function and from there you'll find the sup, inf etc. My first post here on the math section, don't kill me Hope I helped.

Answer 2

The problem is far from being trivial.

The sequence given by $a_n=e^{in}$ is dense in the unit circle since $\pi$ is an irrational number. By considering $\text{Im}(a_n)$, we get that the sequence given by $b_n=\sin(n)$ is dense in the interval $[-1,1]$.

However, that is not enough to prove that the wanted $\sup,\inf,\liminf,\limsup$ are $1,0,0,1$.

By Lagrange's theorem there is an infinite number of rational numbers $\frac{p_n}{q_n}$ such that:

$$ \left| \pi-\frac{p_n}{q_n}\right| \leq \frac{1}{q_n^2} $$ holds, and in such a case: $$ \left|\sin(p_n)\right|\leq \frac{1}{q_n}, $$ but that is not enough to ensure that $\left|\sin(p_n)\right|^{\frac{1}{p_n}}$ is arbitrarily close to zero.

The problem strongly depends on the irrationality measure of $\pi$. Since $\pi$ has a finite irrationality measure (it is conjectured to be $2$, but until today no one has proved something better than $7.6$) our $\inf$ is indeed a minimum and it is strictly greater than zero.

Answer 3

I will show that $\sqrt[n]{|\sin(n)|} \to 1 $ as $n \to \infty $.

Since this depends on how close $n$ can be to $\pi$, what is useful here is the irrationality measure of $\pi$.

It turns out that there is a value $v > 0$ such that, for any rational approximation $\frac{p}{q}$ to $\pi$, $\big|\pi- \frac{p}{q}\big| >\frac1{q^{v}} $. $v = 20$ will work.

Here is one article that shows this: https://projecteuclid.org/download/pdf_1/euclid.pja/1195511637

Therefore, $|n-m\pi| =m|\frac{n}{m}-\pi| > m\frac1{m^{v}} =\frac1{m^{v-1}} $.

Now, we need a bound relating to $n$, not $m$.

Since we can choose $|n-m\pi| < \pi $, $|m\pi| =|n-(n-m\pi)| \le |n|+|n-m\pi| $, or $|n| \ge |m\pi| $, so $\frac1{|n|} \le \frac1{|m\pi|} $, or $\frac1{|m|} \ge \frac{\pi}{|n|} $.

Therefore, $|n-m\pi| >\frac1{m^{v-1}} \ge\frac{\pi^{v-1}}{n^{v-1}} $.

Since $\sin(n) =\sin(n-m\pi) $, and, for $|x| < \frac{pi}{2}$, $|sin(x)| \ge \frac{2x}{\pi} $, $|\sin(n)| =|\sin(n-m\pi)| > |\frac{2(n-m\pi)}{\pi|} \ge |\frac{2}{\pi}\frac{\pi^{v-1}}{n^{v-1}}| = |\frac{2\pi^{v-2}}{n^{v-1}}| $.

Now, we finally get to the conclusion.

From this, $\sqrt[n]{|\sin(n)|} \ge \sqrt[n]{|\frac{2\pi^{v-2}}{n^{v-1}}|} =|\frac{(2\pi^{v-2})^{1/n}}{n^{(v-1)/n}}| $. But, both $a^{1/n} \to 1$ and $n^{1/n} \to 1$ as $n \to \infty$.

Therefore $\sqrt[n]{|\sin(n)|} \to 1 $ as $n \to \infty $.

Note: I am sure this has been done before.