Divergence of {cos(n)}

In Mathematical Spectrum (v 47, 2014/2015, n 2, 68-70) Kurt Fink and Jawad Sadek gave four proofs of divergence of the sequence $\{\cos (n)\}.$ Picking up the gauntlet, I list their proofs and add two more.

In the Introduction to their article, Fink and Sadek remark that many texts dismiss the problem with the remark that cosine is an oscillating function and, for that reason, $\{\cos (n)\}$ can't be convergent. They refute the obvious falsity of this line of thinking with an example of another oscillating function $f(x) = \cos 2\pi x$ for which the sequence $\{f(n)\}=\{\cos (2\pi n)\}$ is as obviously convergent.

So for the proofs ... (Please leave a comment if you see more proofs.)

Observe that convergence of $\{\cos (n)\}$ implies convergence (to the same limit) of $\{\cos (n+2)\}$ and, hence, of the left-hand side in

$\cos (n) - \cos (n+2)=2\sin (n+1)\sin (1),$

implying $\displaystyle\lim_{n\rightarrow\infty}\sin (n+1)=0$ and, thus, $\displaystyle\lim_{n\rightarrow\infty}\sin (n)=0$ But, with the addition formula for sine,

$\sin (n+1) = \sin (n)\cos (1)+\sin (1)\cos (n),$

we arrive at the conclusion that $\cos (n)\rightarrow 0,$ and this - along with $\sin (n)\rightarrow 0$ - contradicts the Pythagorean theorem.

According to de Moivre's formula,

$(\cos (1)+i\sin (1))^{n}=\cos (n)+i\sin (n).$

Since $|\cos (1)+i\sin (1)|=1,$ the point $(\cos (1)+i\sin (1))^{n}$ stays on the unit circle, rotating by 1 radian with the change from $n$ to $n+1$ and, therefore, could not converge. Neither could its components $\cos n$ or $\sin n.$

For $0\lt\beta\pi/2-1,$ define to intervals

$\displaystyle E_{k}=\bigg(2k\pi,(4k\pi +1)\frac{\pi}{2}-\beta\bigg)$ and $\displaystyle O_{k}=\bigg((2k+1)\pi,(4k\pi +3)\frac{\pi}{2}-\beta\bigg).$

The length of this intervals exceeds $1,$ implying that each contains an integer. Denote those $\{e_{k}\}$ and $\{o_{k}\},$ respectively. Now,

$\displaystyle\cos (e_{k})\gt \cos\bigg(\frac{(4k+1)\pi}{2}-\beta\big)=\sin\beta\gt 0$

whereas

$\displaystyle\cos (o_{k})\lt \cos\bigg(\frac{(4k+3)\pi}{2}-\beta\big)=-\sin\beta\lt 0.$

Since the two subsequences $\{\cos (e_{n})\}$ and $\{\cos (o_{n})\}$ are bound way apart, the sequence $\{\cos (n)\}$ could not be possibly convergent.

Let $k$ be an odd positive integer and let $n_{k}$ be such that $\displaystyle n_{k}\lt\frac{k\pi}{2}\lt n_{k}+1.$ By the Mean Value Theorem, there is a $\theta,$ $n_{k}\lt\theta\lt n_{k}+1,$ such that

$\begin{align} |\cos (n_{k}+1)-\cos (n_{k}) &= |\sin (\theta )||(n_{k}+1)-n_{k}|\\ &=|\sin (\theta )|. \end{align}$

This implies that $\displaystyle\theta\in\bigg(\frac{k\pi}{2}-1,\frac{k\pi}{2}+1\bigg),$ so that

$\displaystyle |\sin (\theta )|\in\bigg(\bigg|\sin\bigg(\frac{k\pi}{2}+1\bigg)\bigg|,1\bigg).$

Hence,

$\displaystyle |\cos (n_{k}+1)-\cos (n_{k})|\gt \bigg|\sin\bigg(\frac{k\pi}{2}+1\bigg)\bigg|\gt \bigg|\sin\bigg(\frac{k\pi}{2}+\frac{\pi}{3}\bigg)\bigg| =\frac{1}{2},$

implying that the subsequence $\{\cos (n_{k})\}$ cannot converge, precluding convergence of the whole sequence $\{\cos (n)\}.$

By the double argument formula, $\cos 2x = 2\cos^{2}x-1,$ we also have

$\begin{align} \cos 4x &= 2\cos^{2}2x-1\\ &=2(2\cos^{2}x-1)^{2}-1\\ &=8\cos^{4}x-8\cos^{2}x-1. \end{align}$

Note that if $\{\cos (n)\}$ has limit $L:$ $\displaystyle\lim_{n\rightarrow\infty}\cos (n)=L,$ then also $\displaystyle\lim_{n\rightarrow\infty}\cos (2n)=L$ and $\displaystyle\lim_{n\rightarrow\infty}\cos (4n)=L.$ This gives two equations for $L:$

$L=2L^{2}-1$ and $L=8L^{4}-8L^{2}-1.$

The roots of the first are easy to find with the quadratic formula. The roots of the second could be seen with the modern calculating tools to be nowhere close to the roots of the first equation.

This is an elaboration on the second proof. The idea is that (as it follows from the Pigeonhole principle, the sequence $\{\cos(n)+i\sin (n)\}$ not only divergent, its values form a dense set on the unit circle, implying that the two projections, $\{\cos (n)\}$ on the $x$-axis and $\{\sin (n)\}$ on the $y$-axis are dense in the interval $[-1,1].$ Thus neither can be convergent.