TITLE: Proving that $A=\{(-2)^n : n \in \mathbb{N} \}$ is unbounded
QUESTION [9 upvotes]: I am trying to prove that  $A=\{(-2)^n : n \in \mathbb{N} \}$ is unbounded.
What I did was first to show that for every $n \in \mathbb{N}$ if $n$ is even then $(-2)^n = 2^n$ and  if $n$ is odd then $(-2)^n = -2^n$ (I did it by induction on $n$).
Then I show that for every $n \in \mathbb{N}$ there exist $k \in \mathbb{N}$ such that  $(-2)^n \lt (-2)^k$ because if $n$ is even then
$$(-2)^n=2^n\lt 2^k$$
Where the last inequality holds by the definition of natural powers. The case where $n$ is odd is obvious now.
Am I right? Is the a more elegant way of proving so? thanks!

REPLY [8 votes]: What does it mean to be bounded? $A$ is bounded if there exists some positive number $M$ such that $|a|\leq M$ for all $a\in A.$ You want to show $A$ is unbounded, so that there is no such $M$ that satisfies that condition. One way to do this is to pick an arbitrary positive number, and find something in $A$ which has larger magnitude than what you picked. If you can do that for any $M>0$ then $A$ can't be bounded. 
So to start your proof: Let $M>0.$ We can find $a\in A$ such that $|a|> M$ by picking $n$ ...

REPLY [6 votes]: I would simply prove by induction that $2^n>n$ for $n\in\Bbb N$ and then note that for any $m\in\Bbb N$, $(-2)^{2m}=2^{2m}>2m>m$, so $A$ has no upper bound. If you also want to show that $A$ has no lower bound, just observe that $(-2)^{2m+1}=-2^{2m+1}<-(2m+1)<-m$. (Here I’m using the fact about odd and even powers of $-2$ that you already proved by induction.)