What is $\lim\limits_{k \rightarrow \infty}{(-2)^k}$? What is the $\lim\limits_{k \rightarrow \infty}{(-2)^k}$?
Can this limit even exist? Isn't the function divergent? 
How do I compute $\sum_{k=0}^\infty{(-2)^k}=-2+4-8+16\dots$ ?
For the second part I tried rearranging: $\sum_{k=0}^\infty{(-2)^k}=\sum_{k=0}^\infty{2^{2k-1}}$ I still don't know what this is, first I thought of the geometric series but that only works for $|q|&lt;1$ and 2 obviously isn't smaller than 1. Besides this series is divergent, but the geometric series is konvergent.
If we choose $a_n=2n$ and $b_n=2n+1$, then $a_n\to \infty$ and $b_n\to \infty$. Since 
$$\lim_{n\to \infty}(-2)^{a_n}=\infty,$$
$$\lim_{n\to \infty}(-2)^{b_n}=-\infty,$$
we can conclude that the limit doesn't exist.
The limit does not exist, because the sequence has one subsequence that converges to $\infty$ and another subsequence that converges to $-\infty$.

We also know:


  If $\sum_{n=1}^\infty a_n$ exists, then $\lim_{n\to\infty} a_n=0$.


Therefore, we also know that the series is divergent (because the limit, in this case, is not $0$ - since it does not exist).