On the Divergence of $s_n=\cos{\frac{\pi}{3}n}$: A Proof Question:

Show that $s_n=\cos{\frac{\pi}{3}n}$ is divergent.

Attempt:

Suppose that $\lim_{n\rightarrow \infty}(\cos{\frac{\pi}{3}n})=s$, then given an $\epsilon$, say $\epsilon=1$, we can find an $N\in\mathbb{N}$ so that $$\begin{vmatrix} (\cos{\frac{\pi}{3}n})-s\end{vmatrix}<1.$$
  If $n=6k+1$---for some sufficient $k\in\mathbb{N}$, then we obtain $\lvert \frac{1}{2}-s\rvert<1$, and so $\frac{1}{2}<s<\frac{3}{2}$; however, if $n=6k+3$---likewise for some sufficient $k\in\mathbb{N}$, then we obtain $\lvert -1-s\rvert<1$, and so $-2<s<0$. Therefore, since $s$ cannot satisfy both inequalities, $\lim_{n\rightarrow\infty}(\cos{\frac{\pi}{3}n})$ does not exist


Should look something like this:

Suppose that $\lim_{n\rightarrow \infty}(\cos{\frac{\pi}{3}n})=s$, then given an $\epsilon$, say $\epsilon=\frac{1}{4}$, we can find an $N\in\mathbb{N}$ so that $$\begin{vmatrix} (\cos{\frac{\pi}{3}n})-s\end{vmatrix}<1.$$
  If $n=6k+1$---for some sufficient $k\in\mathbb{N}$, then we obtain $\lvert \frac{1}{2}-s\rvert<\frac{1}{4}$, and so $\frac{1}{4}<s<\frac{3}{4}$; however, if $n=6k+3$---likewise for some sufficient $k\in\mathbb{N}$, then we obtain $\lvert -1-s\rvert<\frac{1}{4}$, and so $-\frac{5}{4}<s<-\frac{3}{4}$. Therefore, since $s$ cannot satisfy both inequalities, $\lim_{n\rightarrow\infty}(\cos{\frac{\pi}{3}n})$ does not exist

As was alluded to by A.N.
