Prove $\lim_{n \to \infty} |\cos (nx)|^{\frac{1}{n} }= 1$ for almost every $x \in \mathbb{R}$

Question

don't know how to proceed with this. my thought: if there exist some $n_0, k_0 \in \mathbb{Z}, x_0 \in \mathbb{R}$ such that $n_0 x_0= \frac{\pi}{2}+k_0\pi $, then there exist countable sequence $\{(n_i,k_i)\}$ such that $n_i x_0= \frac{\pi}{2}+k_i\pi $, so the limit doesn't exist at $x_0$

Answer 1

Let $U\in[0,2\pi)$ be uniformly random. Since $\cos$ has bounded derivative, there exists a constant $C$ such that for every $\delta>0$, $$ \mathbb P(|\cos U|<\delta)<C\delta. $$ Since $nU$ mod $2\pi$ is also uniformly random for all $n\geq 1$, this implies $$ \mathbb P(|\cos nU|<\delta)<C\delta, $$ or equivalently $$ \mathbb P(|\cos nU|^{1/n}<\delta)<C\delta^n. $$ Since for all $\delta<1$ the sum $\sum_{n=1}^{\infty}C\delta^n$ is finite, it follows by the Borel-Cantelli lemma that there are almost surely only finitely many $n$ such that $|\cos nU|^{1/n}<\delta$, and hence that $$ \liminf_{n\geq 1}|\cos nU|^{1/n}\geq \delta\quad \text{almost surely}, $$ for all $\delta<1$. Taking a sequence of $\delta$'s tending to $1$ shows that the liminf is at least $1$ almost surely. But since $|\cos nU|^{1/n}\leq 1$, we get a matching upper bound, showing that the limit exists (almost surely) and equals $1$ (almost surely), as desired.