# The sequence $x^n$

$$\def \C {\mathbb{C}}\def \R {\mathbb{R}}\def \N {\mathbb{N}}\def \Z {\mathbb{Z}}\def \Q {\mathbb{Q}}\def\defn#1{{\bf{#1}}}$$
What about sequences such as $1/2$, $1/4$, $1/8$, $1/16$, which can defined as $(1/2)^n$? The following proposition deals with this. It is very useful and is used countless times in mathematical proofs. The proof here uses properties of the log function, a function we have not defined rigorously but no doubt one you have met before.

Proposition
Suppose that $|x|\lt 1$. Then the sequence $x^n$ has limit $0$.

Proof If $x=0$, then we have a constant sequence and the limit is obviously $0$. Hence, we can assume $x\neq 0$.
Given $\varepsilon >0$, let $N\in \N$ be greater than $\log \varepsilon / \log |x|$.

We have
\begin{eqnarray*}
|x^n-0|\lt \varepsilon &\iff &|x^n|<\varepsilon\\ &\iff & |x|^n \lt \varepsilon \\ &\iff & \log |x|^n \lt \log \varepsilon \\ &\iff & n\log |x| \lt \log \varepsilon \\ &\iff & n > \frac{\log \varepsilon }{\log |x| } .
\end{eqnarray*}
(The inequality sign reverses because $0\lt |x|\lt 1$ means $\log |x|$ is negative.)

Hence, if $n>N$, then $\displaystyle n> \frac{\log \varepsilon }{\log |x| }$ and so by the above we have $|x^n-0|\lt \varepsilon$. That is, $x^n$ has limit $0$.