Show that $\cos(2^n)$ diverges

Question

Problem:

I have sequence defined as follows: $$ a_n = \cos(2^n) $$

I need to show that it diverges.

My progress:

I tried common method here supposing that this sequence has limit $\lim_{n\to\infty}\cos(2^n) = a$ and then trying to get to contradiction with $\lim_{n\to\infty}(\cos(2^{n+1})-\cos(2^n))=0$. But it led me nowhere.

How can I show that this sequence diverges?

That still leaves two possibilities, and there's extra work needed to show neither is the true limit. — Greg Martin, Nov 06 '21 at 16:41
Note that $\cos(2^n \alpha)$ can converge for some $\alpha$, so @nejimban's method will not work. For example $\alpha = 2\pi/3$. — GEdgar, Nov 06 '21 at 16:41
Does this answer your question? What is the cluster set of the sequence ${\cos(2^n)}$? – On a second read, the referenced answer seems to be incomplete. — Martin R, Nov 06 '21 at 16:46
@GEdgar: You can eliminate those possibilities by noting that $\pi$ is irrational, can't you? — Brian Tung, Nov 06 '21 at 16:55
A bit of a funny thought, but your proof must make reference to some property unique to real numbers, otherwise it will be false since I can reproduce the proof that it doesn't converge in the 2-adics where this limit actually converges to 1. — Merosity, Nov 07 '21 at 02:52

score 5 · Accepted Answer · answered Nov 07 '21 at 01:48

$\textbf{Claim:}$ If $a_{n}$ converges then it converges to either $1$ or $-\frac{1}{2}$.

Proof: Suppose it converges to some $c$ then for each $\epsilon > 0$ $\exists N_{\epsilon} \in \mathbb{N}$ so that if $n \geq N_{\epsilon}$ we have

$$|a_{n+1}-a_{n}| < \epsilon$$

$$|2a_{n}^2-a_{n}-1|<\epsilon$$

This means that $c$ lies an arbitrarily small neighbourhood of one of the roots of the polynomial $2x^2-x-1$ which are $1$ and $-\frac{1}{2}$ respectively as we can take $\epsilon \rightarrow 0$.

$\textbf{Claim:}$ $a_{n}$ does not converge to $-\frac{1}{2}$.

Suppose it converges to $-\frac{1}{2}$ then

$$a_{n+1}+\frac{1}{2} = 2a_{n}^2-1+\frac{1}{2} = 2a_{n}^2-\frac{1}{2} = 2(a_{n}+\frac{1}{2})(a_{n}-\frac{1}{2})$$

Let us choose $M \in \mathbb{N}$ so that for all $n \geq M$ we have $|a_{n}-\frac{1}{2}| \geq \frac{3}{4}$. For $n \geq M$ one has $$ |a_{n+1}+\frac{1}{2}| \geq \frac{3}{2}|a_{n}+\frac{1}{2}| $$

One must clearly have $a_{M} = \frac{1}{2}$ because otherwise

$\liminf_{k \rightarrow\infty}|a_{k}+\frac{1}{2}| \geq (\frac{3}{2})^{k-M}|a_{M}+\frac{1}{2}| = \infty.$

But if $a_{M} = \frac{1}{2}$ then $2^{m} \equiv \frac{2\pi}{3} \mod 2\pi$ or $2^{m} \equiv \frac{4\pi}{3} \mod 2\pi$; this would imply that $\pi$ is rational which is impossible. This is a contradiction.

$\textbf{Claim:}$ $a_{n}$ does not converge to $1$.

Suppose it converges to $1$ then

$$a_{n+1}-1 = 2a_{n}^2-2 = 2(a_{n}-1)(a_{n}+1)$$

Let us choose $K \in \mathbb{N}$ so that for $n \geq K$ we have $|a_{n}+1| \geq \frac{3}{2}$.

For $n \geq K$ one has

$$|a_{n+1}-1| \geq 3|a_{n}-1|$$

One must have $a_{K} = 1$ because otherwise we have

$$\liminf_{k \rightarrow \infty}|a_{k}-1| \geq 3^{k-K}|a_{K}-1| = \infty$$

But if $a_{K} = 1$ then $2^{K} \equiv 0 \mod 2\pi$; this implies that $\pi$ is rational. This is a contradiction.

Show that $\cos(2^n)$ diverges

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