Proving $\ln\left(\cos \frac{1}{2^n}\right) = O\left(\frac{1}{4^n}\right)$

Question

I would like to show that :

$$\ln\left(\cos \frac{1}{2^n}\right) = O\left(\frac{1}{4^n}\right)$$

Attempt :

let's show that the limit :

$\lim\limits_{n \to \infty} 4^n\ln\left(\cos \frac{1}{2^n}\right)$ is bounded. Now the problem is that I need to control at which speed $\ln(...)$ go to $0$. Yet I don't se how to do so.

Thank you !

Answer 1

We have $$\cos x=1-\frac{x^2}2+o(x^2)\\\cos\frac1{2^n}=?\\\ln (1-x)=x+o(x)\\\ln\left(\cos\frac1{2^n}\right)=?$$

Can you fill in the blanks?

Answer 2

Consider that, for small $\epsilon$ $$\cos(\epsilon)=1-\frac{\epsilon ^2}{2}+\frac{\epsilon ^4}{24}+O\left(\epsilon ^6\right)$$ Continue with Taylor series to get $$\log (\cos (\epsilon ))=-\frac{\epsilon ^2}{2}-\frac{\epsilon ^4}{12}+O\left(\epsilon ^5\right)$$ Make $\epsilon=\frac 1{2^n}$ and ... continue