How to calculate $\lim_{n\to\infty} \sqrt[n]{|\cos n|}$ I have seen the proof of $\displaystyle\lim_{n\to\infty} \sqrt[n]{|\sin n|}=1$ by using Louivile number. In a similar way , I tried
to prove $\displaystyle\lim_{n\to\infty} \sqrt[n]{|\cos n|}=1$ by using Dirichlet 's approximation theorem:
If $\omega$ is an irrational number ,then  there exist infinite many positive integer $p_n,q_n (q_n>1)$  such that
$$ \left| \omega-\frac{p_n }{q_n }\right|<\frac{1}{q_n^2 }.$$
Take $\omega=\pi$, then
$$ \left| q_n\pi-p_n\right|<\frac{1}{q_n }.$$
Hence $|\cos p_n|=\cos\left| q_n\pi-p_n\right|>\cos\frac{1}{q_n}$.
Therefore we have $\sqrt[n]{|\cos p_n|}>\sqrt[n]{\cos\frac{1}{q_n}}\to 1. (n\to \infty)$
Here the set $\{p_n\}$ is just a part of the natural number set $\mathbb{N}.$  I donot know how to  go further.
I would appreciate if someone could give some suggestions and comments. Welcome new proof. Thanks in advance.
 A: Dirichlet approximation goes in the wrong direction.  You need the fact that rational approximations of $\pi$
can't be too good, not the fact that there are approximations that are not too bad.
In order to have $|\cos(n)|^{1/n} < 1-\epsilon$
(for some constant $\epsilon \in (0,1)$) we would need
$|\cos(n)| < (1-\epsilon)^n$, which would imply
$|n - k\pi/2| < 2 (1-\epsilon)^n$ for some odd $k$,
and thus $|\pi - 2n/k| < 4 (1-\epsilon)^n/k$.  Now it is known that $\pi$ has finite irrationality measure: the
best bound at the moment  (Zeilberger and Zudlin 2019) on the irrationality measure of $\pi$ is that it is at most $7.103205334137$.  That is, for any $\mu$ greater than that number, there are at most finitely many pairs of positive integers $p,q$ with
$$ \left| \pi - \frac{p}{q} \right| < \frac{1}{q^\mu} $$
We want to take $p=2n$, $q=k$, and note that $(1-\epsilon)^n/k < 1/k^\mu$ for sufficiently large $n$ and $k$ when $2n/k$ is near $\pi$.  The conclusion is that $|\cos(n)|^{1/n} \ge 1-\epsilon$ for sufficiently large $n$, and thus $|\cos(n)|^{1/n} \to 1$ as $n \to \infty$.
