How does one prove by definition that $a_n=\cos(n^2\pi) $ converges as $n$ approaches infinity?

Question

I've studied the definitions of convergence and divergence, but I still cant see how to approaches these types of problems, I know that the sequences oscillates between $1$ and $-1$ but I can't show that using mathematical definitions, I thought of picking $2$ sub-sequences and showing that their limit is different and that implies that my original sequence diverges, any tips on how to pick these sequences, if this even solves the problem at all.

Answer 1

The sequence diverges because $a_{2n} = 1, a_{2n+1} = -1$ for all naturals $n$.

Answer 2

Assuming that $a_{n}\rightarrow L$ for some $L$, then we may find some $N\in{\bf{N}}$ such that if $n\geq N$, then $|a_{n}-L|=|\cos(n^{2}\pi)-L|<1$. In particular we have $|\cos(2N)^{2}\pi-L|=|L-1|<1$ and that $|\cos(2N+1)^{2}\pi-L|=|L+1|<1$, then $2=|(L+1)-(L-1)|\leq|L+1|+|L-1|<2$, a contradiction.