Any suggestions on how to compute $\limsup |\cos n|^{n^2}$? This problem has proven very difficult, does anyone have any suggestions on how to tackle it? Any little known theorems/identities that might help?
 A: Let us consider the continued fraction of $\pi$:
$$ \pi = [3; 7, 15, 1, 292, 1, 1, 1, 2, 1, 3, 1, 14,\ldots]$$
together with its convergents $\frac{p_1}{q_1}=\frac{22}{7},\frac{p_2}{q_2}=\frac{333}{106},\frac{p_3}{q_3}=\frac{355}{113}$ etcetera.
The sequence $\{p_n\}_{n\geq 1}$ provides a set of natural numbers where the cosine function attains values estremely close to $\pm 1$. By the properties of continued fractions we have $\left|\frac{p_n}{q_n}-\pi\right|\leq \frac{1}{q_n^2}$, hence $\left|p_n-\pi q_n\right|\leq \frac{1}{q_n}\approx\frac{\pi}{p_n}$ and by the Lipshitz-continuity of $\sin$
$$\left|\cos(p_n)\right|^{p_n^2}=\left(1-\sin^2 p_n\right)^{p_n^2/2} \approx \left(1-\frac{\pi^2}{p_n^2}\right)^{p_n^2/2}\to e^{-\pi^2/\color{red}{2}}\geq \frac{1}{140}$$
so the wanted $\limsup$ is at least as large as $\frac{1}{140}$. On the other hand, if we assume that the terms of the continued fraction of $\pi$ are unbounded, the constant $2$ above can be replaced by any larger constant and the wanted $\limsup$ is just one. The pity is we do not actually know if the terms of the continued fraction of $\pi$ are unbounded or not. For sure, if we consider $n=710$ the value of $\left|\cos n\right|^{n^2}$ is very close to $1$, being equal to $\approx 0.999084$. Additionally, since $\pi\not\in\mathbb{Q}(\sqrt{5})$, infinite terms of the continued fraction of $\pi$ are $\geq 2$, from which we have WimC's claim
$$ \limsup_{n\in\mathbb{N}}\left|\cos(n)\right|^{n^2}\geq e^{-\pi^2/8}>\frac{23}{79}.$$
Following the same lines we have
$$ \limsup_{n\in\mathbb{N}}\left|\cos\frac{\pi n}{e}\right|^{n^2}=1$$
since the terms of the continued fraction of $e$ are unbounded, due to
$$ e=[2;1,2,1,1,4,1,1,6,1,1,8,1,1,10,\ldots].$$
