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

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