Proof of divergence of $\{cos(\frac{n\pi}{2}))\}$

Question

How would I show that the sequence $\{\cos(\frac{n\pi}{2}))\}$ diverges using an even/odd argument? So, here's what I have so far: \

Suppose $\{\cos(\frac{n\pi}{2}))\}$ converges to $L \geq 0$. Choose $\epsilon = 1/2$. Then, $\exists N$ such that $\forall n > N$ and $n \equiv 2 \pmod 4 $ we get $\lvert -1-L \rvert = 1+L \geq 1 > 1/2$ which means it can't converge to $L \geq 0$.
Suppose $\{\cos(\frac{n\pi}{2}))\}$ converges to $L < 0$. Choose $\epsilon = 1/2$. Then, $\exists N$ such that $\forall n > N$ and $n \equiv 0 \pmod 4 $ we get $\lvert 1-L \rvert = 1+\lvert L \rvert >1 > 1/2$ which means it can't converge to $L < 0$.

Now, since $\cos(\frac{n\pi}{2})$ takes on values $1,0,-1,0$ based on $n = 1,2,3,4 ...$ , do I need to show for the case where $n \equiv 1,3 \pmod 4$? I feel that I am missing something in this proof.

Answer 1

I would show divergence by stating two different subsequences that converge to different limits, which implies that the original sequence diverges. formally, if

$a_{n}=\cos \frac{\pi n}{2}$

so take $\left\{ a_{4k}\right\} _{k=1}^{\infty}$ and $\left\{ a_{4k+2}\right\} _{k=1}^{\infty}$ which converge to 1 and -1 respectively

Answer 2

Alternatively, it isn't Cauchy. In particular, for any $N$, let $m=4N, n=4N+2$. Then $n,m\ge N$ and $|a_n-a_m|=|-1-1|=2$.