Does the limit of $\cos^{2n}(n)$, with $n$ a positive integer, converges as $n\to\infty$? I'm struggling with what it seems to be a pretty simple limit:
$$\lim_{n \rightarrow \infty} \cos^{2n}(n)$$
I have arguments to believe that this limit converges to $0$ because $n \in (kπ, (k+1)π) $ and $\cos[(k\pi, (k+1)\pi)] \rightarrow 0 $ as $n$ increases. But also I believe that this limit may diverge, because you can always find an integer that is closer to a multiple of $\pi$ (by Dirichlet's approximation theorem), so you may find a subsequence that converges to 1, and so this limit diverges.
Many thanks in advance!!
 A: By considering the convergents of the continued fraction of $\pi$ we get infinite rational numbers $\frac{p_n}{q_n}$ such that $|p_n-\pi q_n|\leq \frac{1}{q_n}$. Since $\cos(x)=-1+\frac{1}{2}(x-\pi)^2+o(x-\pi)^2$  it follows that $\limsup\cos(n)^{2n}=1$. On the other hand an infinite number of $p_n$s is even. In such a case, by considering how close $\cos(p_n/2)$ is to zero, we get $\liminf\cos(n)^{2n}=0$, so the wanted limit does not exist.
A: I will (shamefully) admit that I am not too proficient in using O-type notation.
Here's a rather extensive version of Jack's proof.
Consider the convergents $p_n/q_n$ of the continued fraction for $\pi$ and bear in mind that
$$\lim_{n\to\infty}q_n = +\infty \tag{1}$$
$$|p_n-q_n\pi|\leqslant \frac1{q_n}\tag{2}$$
By Taylor's Theorem with Lagrange remainder, we have that
$$\cos(x) = \cos(a) - \cos(a)\,\frac{{(x-a)}^2}2+R_a(x),\tag{3}$$
where $R_a(x) = \frac{\sin(\eta)}{3!}\,{|x-a|}^3$ for some $\eta$ between $a$ and $x$.
In particular,
$$|R_a(x)| \leqslant \frac{{|x-a|}^3}{6}\tag{4}.$$
Inputting $x=p_n$ and $a=q_n\pi$ into equation $(3)$ we obtain
$$\cos(p_n) = 1 - \frac12{(p_n-q_n\pi)}^2+R_{q_n\pi}(p_n),\tag{5}$$
where via $(4)$ we have the bound
$$|R_{q_n\pi}(p_n)| \leqslant \frac16 {|p_n-q_n\pi|}^3\leqslant \frac1{6q_n^3}\tag{6}.$$
In particular, as $n\to\infty$ it follows from $(1)$ that $|R_{q_n\pi}(p_n)| \to 0$.
We will consider the expression ${\cos(p_n)}^{2p_n} = \exp\Big(2p_n\,\log\big(\cos(p_n)\big)\Big)$.
We once again apply Taylor's Theorem with Lagrange remainder, to obtain that
$$\log(1-x) = -x + \mathcal R_1(1-x),\tag{7}$$
where $\mathcal R_1(1-x) = -\frac12 {\left(\frac x\xi\right)}^2$ for some $\xi$ between $1$ and $1-x$.
We can hence write
\begin{align}\log(\cos(p_n))
&=
\log\Big( 1 - \left(\frac12{(p_n-q_n\pi)}^2-R_{q_n\pi}(p_n)\right)\Big)
\\&=
R_{q_n\pi}(p_n) - \frac12{(p_n-q_n\pi)}^2 - \frac12 {\left(\frac{R_{q_n\pi}(p_n) - \frac12{(p_n-q_n\pi)}^2}\xi\right)}^2
\end{align}
so that
\begin{align}
|2p_n\,\log(\cos(p_n))|
&\leqslant
\underbrace{2p_n\left\lvert R_{q_n\pi}(p_n)\right\rvert}_{\displaystyle\leqslant \frac{p_n}{3q_n^3}}
+
\underbrace{p_n{(p_n-q_n\pi)}^2}_{\displaystyle\leqslant \frac{p_n}{q_n^2}}
\\&+\underbrace{\frac1{\xi^2}}_{\xi\to 1 \,\text{ as }n \to \infty}\left(
\underbrace{p_n{\left\lvert R_{q_n\pi}(p_n)\right\rvert}^2}_{\displaystyle\leqslant \frac{p_n}{36q_n^6}}
+\underbrace{p_n\left\lvert R_{q_n\pi}(p_n)\right\rvert(p_n-q_n\pi)}_{\displaystyle\leqslant \frac{p_n}{6q_n^4}}
+\underbrace{\frac{p_n}2{(p_n-q_n\pi)}^4}_{\displaystyle\leqslant \frac{p_n}{2q_n^4}}
\right).
\end{align}
Because of $(1)$ and $\lim_{n\to\infty} p_n/q_n = \pi$, it follows that whenever $a>0$ we have
$$\lim_{n\to\infty} \frac{p_n}{q_n^{1+a}} = \lim_{n\to\infty} \frac{p_n}{q_n}\,\frac1{q_n^a} = 0.$$
Hence, $\lim_{n\to\infty} 2p_n\,\log(\cos(p_n)) = 0$ and finally
$$\lim_{n\to\infty}{\cos(p_n)}^{2p_n} = \lim_{n\to\infty}\exp\Big(2p_n\,\log\big(\cos(p_n)\big)\Big) = \exp(0) = 1.$$
