For $a\in \mathbb{R}$: Does $a^n \rightarrow 0$ when $|a| < 1$ imply that $a^n$ is unbounded when $|a| > 1$?

Question

On pages 50-51 in the book Introduction to Analysis by Rosenlicht, the author claims the following (see yellow highlight):

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How does this readily follow as a consequence of the example?

I don't think the example is sufficient to conclude that $a^n$ grows without bound when $|a| > 1$, there's a detail missing.

For example we need to use some form of Archimedean property to show that for a given real $M$ there exists an integer $m \geq 1$ such that $a^m > M$). Or use properties of exponentials by writing $|a|^n = e^{ n \cdot \ln |a| }$, whose behaviour we are familiar with.

And such a detail makes no reference to the fact that $a^n \rightarrow 0$ when $|a| < 1$. So how exactly is the unbounded case a consequence of the bounded case? Am I missing something obvious here?

Answer 1

If $|a| > 1$, then $|1/a| = 1/|a| < 1$ and consequently by what is already established $$\lim_n \frac{1}{|a|^n}= 0$$ which is equivalent with $$\lim_n |a|^n = \infty$$

Hence, $(a^n)_{n=1}^\infty$ is unbounded.

Answer 2

Suppose that $|a|>1$ and that $M>0$. Since $\lim_{n\to\infty}\left(\frac1a\right)^n=0$, if $n$ is large enough, $\left|\frac1{a^n}\right|<\frac1M$. But$$\left|\frac1{a^n}\right|<\frac1M\iff|a^n|>M.$$This proves that $(a^n)_{n\in\Bbb N}$ is unbounded.