Negative 1 to the power of Infinity

If a sequence converges all its subsequences converge to the same limit.

Note that $(-1)^{2n}$ is a subsequence that converges to $1$ and $(-1)^{2n + 1}$ a subsequence that converges to $-1$. Contradiction.

This does not exist. If you have a sequence $\{x_n\}$, then if $x_n \rightarrow l$, for any open interval $I$ with $l\in I$, $x_n\in I$ for all but finitely many $n$. Intuitively, any open interval containing the limit must "eventually absorb" the sequence.

Your sequence has no such behavior. If you take the interval $(.9, 1.1)$, we hve $x_n\not\in I$ if $n$ is odd. Likewise, taking a small interval around $-1$ results in $x_n$ failing to be in that interval if $n$ is even. There is no point you can pick to eventually absorb the sequence, and therefore there is no limit.

A limit of a sequence exists when there is a number $L$ such that for every $\epsilon>0$ there is some $N\in\mathbb N$ such that for every $n>N$ we have $|a_n-L|<\epsilon$.

The limit is infinite if for every $M>0$ there is some $N\in\mathbb N$ such that for all $n>N$ we have $a_n>M$.

In this case $a_n=(-1)^n$, which takes two values: $1,-1$.

We do not even need our counterexample $\epsilon$ to be small, just set $\epsilon=1$. Then for every number $L$ we have:

If $|1-L|<1$ then $|-1-L|\ge 1$, and if $|-1-L|<1$ then $|1-L|\ge 1$.

Therefore for every $N\in\mathbb N$ either $a_{N+1}$ or $a_{N+2}$ is of at least distance of $1$ from $L$, for any given $L$.

The limit cannot be infinite either, for obvious reasons.

We are therefore left only with the possibility that the limit does not exist.