How to prove that Limit $\lim_{n\to\infty} \cos(n\pi)$ doesn´t exist, or exist?

Question

I believe that when n tends to infinity, the expression $\cos(n\pi)$ doesn´t converge, because the oscilation values. But, how can we formally prove that? Thank you!!

Answer 1

If $n$ is even $cos(n\pi) = 1$, and if $n$ is odd $cos(n\pi) = -1$. So, there is a subsequence ($cos(2n\pi)$) that converges to $1$ and another subsequence ($cos((2n+1)\pi)$) that converges to $-1$. So, the sequence does not converge.

Answer 2

Just do the definition.

If $\lim_{n\to \infty} \cos (n\pi) = a$ then any $\epsilon > 0$ the there is an $N$ so that if $n > N$ then $|\cos(n\pi)-a| < \epsilon$.

Well if $n$ is even than $\cos (n\pi) = 1$ and if $n$ is odd then $\cos (n\pi)=-1$.

So both $|1- a| < \epsilon$ and $|-1 -a | < \epsilon$ which can be shown to be impossible for $\epsilon \le 1$ by the triangle inequality.

$2= |-1-1|=|(-1-a) +(a-1)|\le |-1 + a|+|a-1|=|-1+a|+|1-a|< \epsilon + \epsilon = 2\epsilon < 2$.

A contradiction. so $\lim_{n\to \infty}\cos(n\pi) \ne a$ for and $a\in \mathbb R$.

I'll leave it to you to show $\lim_{n\to \infty}\cos(n\pi) \ne \pm \infty$.

($\lim_{n\to \infty}\cos(n\pi) =\infty$ would imply for any $M\in \mathbb R$ there is an $N$ so that $n > N$ would mean $\cos(n\pi) >M$ but $\cos (n\pi) \le 1$.)

Answer 3

If the limit exists, then the sequence $(x_n)$ given by $x_n:=\cos(\pi n), n\in\Bbb N$ converges as well. The sequence isn't Cauchy though, because of the oscillation. Successive terms differ by at least $2$. More precisely, given $\epsilon\lt2$, there is no $N$ such that $n,m\ge N\implies|x_n-x_m|\lt\epsilon$.