What is $\lim\limits_{k \rightarrow \infty}{(-2)^k}$?

Question

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 $|q|<1$ and 2 obviously isn't smaller than 1. Besides this series is divergent, but the geometric series is konvergent.

Answer 1

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.

Answer 2

If you work in the real numbers the other two answers are completely correct. If you work over de $2$-adic numbers $\mathbb{Q}_2$ (the special case of $p$-adic numbers (https://en.wikipedia.org/wiki/P-adic_number) where $p = 2$) then the sum does converge and exactly for the reason you state: in the $2$-adic world we DO have that $|-2| < 1$. (Where the left hand side is usually written $|-2|_2$ to distinquish it from the ordinary absolute value on $\mathbb{R}$.)

You can use the formula for geometric series to find that $2$-adically your series adds up to $1/3$. Your nicer reformulation of the series as $\sum 2^{2k+1}$ then gives you an answer to what the $2$-adic (as opposed to decimal) expansion of $1/3$ looks like. (See the Wikipedia article for more on this latter topic.)

Answer 3

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).