Epsilon - N Proof of a diverging sequence to infinity

Question

I want to prove the sequence converges or diverges using an $\epsilon-\mathit N$ proof.

$a_n= (-2)^n$

I'm really confused because firstly it's been a while since I did these $\epsilon-\mathit N$ proofs and secondly I don't know whether I should be proving by contradiction or doing a limit to infinity proof? I know this diverges where if n is even then you get $\infty$ and if n is odd you get $-\infty$.

Answer 1

Prove by contradiction. Suppose $a_n \to l$. Then there exists $n_0$ such that $|(-2)^{n}-l| <1$ for all $n \geq n_0$. This gives $|(-2)^{n}| =|(-2)^{n}-l)|+l < 1+|l|$ for all $n \geq n_0$. Arrive at a contradiction by taking $n$ such that $n >\log_2(1+|l|)$ and $n \geq n_0$.