Proving that $\{ (-2)^n : n \in \mathbb{N} \}$ is unbounded How do I go about proving that the set $A=\{ (-2)^n : n \in \mathbb{N} \}$ is unbounded?
I started by showing (using induction) that if $n$ is even then $(-2)^n=2^n$, and since $\mathbb{N}$ is unbounded in $\mathbb{R}$ oubviously $A$ is unbounded. Am I in the right direction? Is it enought to say that for every $n \in \mathbb{N}$ there exist $k \in \mathbb{N}$ such that $(-2)^n=2^n \lt 2^k=(-2)^k$?
If I'm write then showing that $A$ is unbounded from beloew is almot similar.
Any thoughts? Thanks!
 A: The argument in your first sentence is correct. But you need to justify the statement "obviously $A$ is unbounded". Why is this so? Well, of course, because $A$ contains all even powers of $2$ and the set of all even powers of 2 is unbounded from above.
So,
what  you need to show is  that: 

given any positive integer $n$, there is an even positive integer $k$
  so that $2^{ k}\ge n$.

The above claim  would follow from a more general statement: 

given any positive integer $n$, the inequality $2^n\ge n$ holds.

You can prove this statement by induction.

So, I would start of by proving the Lemma: $2^n\ge n$ for any positive integer $n$.
Then to show $A$ is unbounded, argue as follows:  
Let $n$ be a positive integer. 
If $n$ is even, then  $2^n=(-2)^n$ is in $A$ and from the lemma we have  $(-2)^n=2^n\ge n$.
If $n$ is odd, we have $2^{n+1}=(-2)^{n+1}$ is in $A$ and, again from the lemma,  $(-2)^{n+1}=2^{n+1}\ge n+1>n$.
So in either case, we have an element in $A$ greater than or equal to $n$.  Since $n$ was an arbitrary positive integer, it follows that $A$ is not bounded from above.

Note that, to show that your $A$ is not bounded, you need only show that $A$ is not bounded from above (or show that it is not bounded from below). Of course, if you were required to show that $A$ is unbounded in  both directions, you'd use a similar argument to show that $A$ is not bounded from below.
