# 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.

• Use $\pi$ for $\pi$, $\cos$ for $\cos$, and $\lim$ for $\lim$. – Shaun Aug 28 '18 at 20:21
• To supplement @Shaun’s comment: here’s the complete MathJax tutorial. – gen-ℤ ready to perish Aug 28 '18 at 22:29
• Limits don't converge or move around; they either exist or don't. – zhw. Aug 28 '18 at 22:49

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.

• Thanks for your response, but still cannot see the connection between what is before the "it follows" and the conclusion after "it follows". Could you please develop a little more your reasoning, please? – Carlos Toscano-Ochoa Aug 29 '18 at 12:26
• Maybe you can help me in this other problem, which is directly related: math.stackexchange.com/questions/2898385/… – Carlos Toscano-Ochoa Aug 29 '18 at 14:53
• @CarlosToscano-Ochoa: the meaning of "it follows" is: for any positive constant $C<1$, $$\lim_{n\to +\infty}\left(1-\frac{C}{n^2}\right)^n = 1,\qquad \lim_{n\to +\infty}\left(\frac{C}{n}\right)^n=0.$$ – Jack D'Aurizio Aug 29 '18 at 19:40
• I must be missing something; I don't really follow either. Is one supposed to take $x=p_n$? – Fimpellizieri Aug 30 '18 at 13:56
• @Fimpellizieri: of course, one is supposed to take some $n$ which is (very) close to an integer multiple of $\pi$ or to an odd multiple of $\pi/2$. – Jack D'Aurizio Aug 30 '18 at 14:21

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

$$\lim_{n\to\infty}{\cos(p_n)}^{2p_n} = \lim_{n\to\infty}\exp\Big(2p_n\,\log\big(\cos(p_n)\big)\Big) = \exp(0) = 1.$$