I have to prove that the sequence $\{ \cos (\frac{n \pi}{3}) \}$ does not converge.
My attempt:
Assume that the sequence $\{ \cos (\frac{n \pi}{3} ) \}$ converges to a real number $l$. In notational terms, $$ \lim \cos (\frac{n\pi}{3} ) = l$$
By the definition, we can make the difference between $l$ and $ \cos (\frac{n \pi}{3})$ as little as we wish to. So, let's make the difference between them less than $1$. There exists an $N$ such that $$ n \gt N \implies \big| \cos ( \frac{ n \pi}{3} ) - l \big | \lt 1$$
for an $n$ which is multiple of $3$, odd and greater than $N$, we have \begin{align} \big| \cos ( m\pi) - l \big| & \lt 1 ~~~~n/3 = m\;\text{ which is odd}\\ | l +1| &\lt 1 \tag{1} \end{align} for an $n$ which is multiple of $3$, even and greater than $N$, we have \begin{align} \big| \cos (k \pi) - l \big| & \lt 1 ~~~~~~~~~n/3 = k\;\text{ which is even}\\ |1-l| & \lt 1 \tag{2} \end{align} Case 1: Consider $l$ to be positive. If $l$ is positive, then $(1)$ cannot be true. Thus, the contradiction is reached.
Case 2: Consider $l$ to be negative. If $l$ is negative, then $(2)$ cannot be true. Thus, the contradiction is reached.
Case 3: Consider $l = 0$. If $l= 0$, then both $(1)$ and $(2)$ are false. Thus, the contradiction.
We have reached the contradiction, because our assumption was wrong about the convergence of the sequence $\cos ( \frac{ n \pi}{3})$. Thus, the sequence $\cos ( \frac{ n \pi}{3})$ does not converge.
Is my solution correct and formal? Formal in the sense that is it in the language of scientific journals?
Yes, I think that is okay. But usually the convergence of this type of sequences are tend to be analyzed with the convergence of its subsequences. For example:
The subsequence $\big( \cos \big( \frac{(6k+3) \pi}{3} \big) \big)_{k \in \mathbb{N}}$ is constantly $-1$ and the subsequence $\big( \cos \big( \frac{(6k) \pi}{3} \big) \big)_{k \in \mathbb{N}}$ is constantly $1$. So the limit cannot exist.