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


You are correct in your work.
However, as suggested above, an easier way is to just show that there are two subsequences converging to different limits.
In your case, $s_{6n+1}$ converges to $0.5$ and $s_{6n+3}$ converges to $-1$.
