Prove that $a_n=(-1)^n$ does not converge

I've managed to come up with a proof to this problem but I don't know if it's right.

Prove that the sequence $a_n=(-1)^n$ does not converge.

Suppose the opposite, that

$(\forall \epsilon > 0)(\exists n(\epsilon) \in \mathbb{N})$ such that $\forall n(\epsilon) > N$ we have:

$|(-1)^n-a|<\epsilon$.

Set $\epsilon=1$, then

$|(-1)^n-a|<1$

Note that for $n$-even we have:

$|(-1)^n-a|=|1-a|<1\\ \Leftrightarrow -1<1-a<1\\ \Leftrightarrow -2<-a<0\\ \Rightarrow -a\in(-2,0) $

And for $n$-odd we have:

$|(-1)^n-a|=|-1-a|<1\\ \Leftrightarrow -1<-1-a<1\\ \Leftrightarrow 0<-a<2\\ \Rightarrow -a\in(0,2) $

So, $-a\in(-2,0)\cap(0,2)\Rightarrow -a\in \emptyset$. We have concluded that the limit $a$ does not exist therefore the sequence $a_n=(-1)^n$ does not converge.

Solution 1:

Yes, your proof by contradiction is correct. You assume that the limit $a$ exists and reach a contradiction that $-a\in\emptyset$.

Alternatively, you can simply observe that for all $n$: $$ |a_{2n}-a_{2n+1}|=2 $$ So the sequence cannot be Cauchy and thus is not convergent.