If $x\in(0,1)$ show the sequence $(y_n)=x^n$ is monotone.

Question

If $x\in(0,1)$ show the sequence $(y_n)=x^n$ is monotone.

Every time I think I'm starting to understand how to do proofs I stumble across a simple problem that stumps me. I could do this very easily if $x \notin (0,1)$, that is, if $x$ was almost any other number. But the simple fact that the series is contracting is throwing me off.

Should I use induction here? Or can I simply say $x^n \geq x^{n+1}$. But then I get $log_{x} x^n \geq log_{x}x^{n+1}$ which implies $n \geq n +1$. Which gives me the contradiction $0 \geq 1$.

Frustrated. Any help would is appreciated.

Answer 1

$$0 < \dfrac{x^{n+1}}{x^{n}}=x<1$$ $$0 < \dfrac{x^{n+1}}{x^{n}}<1 \implies 0<x^{n+1}<x^{n}$$

In the last step I multiplied by $x^{n}$ using $0<x<1$.

Answer 2

$$x^n>x^{n+1}$$ it's $$1>x,$$ which is true.

Thus, $y$ is a decreasing sequence.

Answer 3

I'm curious how you'd handle the case where $x \notin (0,1)$ because to me that seems more complicated (and not always true that $\{x^n\}_n$ is monotone for $x \notin (0,1)$).

Here's one way to use contradiction, which is probably how I'd do it:

Suppose that $\{ x^n \}_n$ were not monotone decreasing. Then there exists some $n$ such that $x^n \le x^{n+1}$.

Since $x > 0$, then $x^n$ is never $0$ for any $n$. So it's safe to divide both sides by $x^n$. Then you get $1 \le x,$ which contradicts $x \in (0,1)$.