Does this sequence $\,\sqrt[n]{1+\cos2n}\,$ have a limit?

Question

Does this $\,\lim_{n\to\infty}\sqrt[n]{1+\cos2n}\,$ exist?

It took me and my tutor 2 days to find this limit. We believe that this function has no limit or if it does, we are not intellectual enough to find it. Hope someone can help us.

Answer 1

Answer: $$ \sqrt[n]{1+\cos(2n)}\to 1. $$

This is due to the fact that the irrationality measure $\mu$ of $\pi$ is a finite number (approx. 7.1032), i.e., for every $\varepsilon>0$, $$ \left|\pi -\frac{p}{q}\right|\ge \frac{1}{q^{\mu+\varepsilon}}, $$ for all but finitely many rationals $p/q$.

This implies that if $2n$ is very close to some $(2k+1)\pi$, where $k\in\mathbb N$, then $$ 1+\cos(2n)=2\cos^2(n)=2\sin^2\big(n-(k+\tfrac{1}{2})\pi\big) \ge \frac{8}{\pi^2}\big(n-(k+\tfrac{1}{2})\pi\big)^2 \\ =\frac{2(2k+1)^2}{\pi^2}\Big(\frac{2n}{2k+1}-\pi\Big)^2\ge \frac{2}{\pi(2k+1)^{2(\mu-1+\varepsilon)}} \approx \frac{c}{n^{2(\mu-1+\varepsilon)}} $$ for some $c>0$. Note that the first inequality above is due to the fact that $$ \sin x>\frac{2x}{\pi}, \quad x\in (0,\tfrac{\pi}{2}). $$ So $$ \sqrt[n]{2}\ge \sqrt[n]{1+\cos(2n)}\ge \sqrt[n]{\frac{c}{n^{2(\mu-1+\varepsilon)}}} $$ and hence $\sqrt[n]{1+\cos(2n)}\to 1$.