Prove that the set $\{a^n:n\in\mathbb{N}\}$ is not bounded above

Question

Fix $a>1$. Prove that the set $S=\{a^n:n\in\mathbb{N}\}$ is not bounded above.

We are given the hint: "First find a positive integer $n$ such that $a>1+\frac{1}{n}$ and prove that $a^n>\left(1+\frac{1}{n}\right)^n\geq2$." However, I do not see how this is helpful.

Attempt: Assume that $S$ is bounded above. Then by the Axiom of Completeness, $S$ has a least upper bound $\sup S$ satisfying, for all $n\in\mathbb{N}$, $\sup S\geq a^n>1$. Then we have that $\sup S-1>0$, and by the Archimedean Property there exists some $N\in\mathbb{N}$ such that $\sup S>\frac{N+1}{N}$.

I have no idea where to go from here or if I am even on the right track. I do not want a full answer, only a hint - either one in the direction of the original hint or one that takes a different approach.

Answer 1

Okay, the hint is that for any $a > 1$ you can find an $n$ so that $a > 1 + \frac 1n$. (Prove that can be done)

That would mean that $(1+\frac 1n)^n \ge 2$. (Prove that) and so $a^n > (1+\frac 1n)^n\ge 2$.

So $a^{kn} \ge 2^k$.

So if you can prove that $\{2^k\}$ is unbounded you'd be done, wouldn't you? For any $M$ there will exist an a $k$ so that $M < 2^k < a^{kn}$.

Answer 2

Suppose $S=\lbrace a^n:n\in \mathbf{N} \rbrace $ is bounded above. Then by the Axiom of Completeness, S has a least upper bound $sup S$ satisfying, $a^n\leq supS$, for all $n\in \mathbf{N}$. Then $a^{n+1}\leq supS \Rightarrow a^n\leq \frac{supS}{a}, \forall n\in \mathbf{N}$, which gives a contradiction since $\frac{supS}{a} < supS$ as $a>1$.

Answer 3

Your hint does require some further effort as mentioned in answer from fleablood.

A much simpler approach is to note that $$a^n=(1+b) ^n>nb$$ where $b=a-1>0$. And clearly the result follows if one notices the obvious fact that sequence $nb$ is unbounded.