Prove that the sequence $\cos(n\pi/3)$ does not converge

Question

EDIT: Using a rigorous formal proof, I need to prove that this sequence does not diverge. I of course understand why it doesn't converge...

$n=1$ to infinity of course.

So, I have a bit of trouble understanding the definition. Here's what I have so far:

I know it doesn't converge if there exists an $\varepsilon > 0$, where for ALL natural numbers $\mathbb{N}$, $|a_n - L| < \varepsilon$, for some $n\in\mathbb N$.

So I can make $\varepsilon > 1/2$, but then I'm not sure what to do. I know I need to prove this by way of contradiction using 2 cases, one where I evaluate $L$ as both greater than and equal the other where it's less than zero.

Answer 1

We give a formal "$\epsilon$-$N$" proof. Note that it is more clumsy and tedious than the other answers that, instead of going back to the definition, use properties of convergent sequences.

Note that if $n$ is $3$ times an even number $2k$, then $\cos(\pi n/3)=\cos(2k\pi)=1$.

Note also that if $n$ is $3$ times an odd number $2k+1$, then $\cos(\pi n/3)=\cos((2k+1)\pi)=-1$.

Suppose that our limit exists and is equal to $a$. Let $\epsilon=1/10$. We show that there is no $N$ such that if $n\gt N$ then $|\cos(\pi n/3)-a|\lt \epsilon$

Suppose to the contrary that there is such an $N$. There is an $n\gt N$ of the form $(3)(2k)$. From this we conclude that $|1-a|\lt \epsilon$.

Similarly, we conclude that $|-1-a|\lt \epsilon$. So $|1+a|\lt \epsilon$.

We have $2=(1-a)+(1+a)$, and therefore by the Triaangle Inequality $$2\le |1-a|+|1+a|\lt 2\epsilon=\frac{1}{5},$$ which is impossible.

Remark: The Triangle Inequality stuff has a mysterious air. But really all that part of the argument does is to say that $a$ cannot be simultaneously within $\frac{1}{10}$ of $1$ and within $\frac{1}{10}$ of $-1$. We could, by the way, have picked $\epsilon=1$.

Edit: We can alternately end by supposing, as in your post, that the limit is $L$. We show that $L$ can be neither $\le 0$ nor $\gt 0$. We show it cannot be $\gt 0$, and leave the other to you. Let $\epsilon=\frac{1}{10}$ and let $N$ have the property that if $n\gt N$ then $\cos(\pi n/3)$ is within $\epsilon$ of $L$. We can find an $n=(3)(2k+1)\gt N$. For that $n$ we have $\cos(n \pi/3)=-1$. But $|-1-L|\gt 1$, contradicting the fact it is $\lt \epsilon$.

Answer 2

HINT: The sequence is periodic (and not constant), hence the limit cannot exist.

Answer 3

If the sequence converges, then all its subsequences have the same limit. However, the subsequence $\big( \cos \big( \frac{(6k+3) \pi}{3} \big) \big)_{k \in \mathbb{N}}$ converges to $-1$ and the subsequence $\big( \cos \big( \frac{(6k) \pi}{3} \big) \big)_{k \in \mathbb{N}}$ converges to $1$. Hence, the limit does not exist.