Show that $\limsup \frac1{x_n}\cdot \limsup x_n\geq 1$

Question

Given a sequence $x_{n}$ and initial data that $0< a\leq x_{n}\leq b< \infty $, for $a,b\in \mathbb{R}$.

I need to show that: $$\limsup \frac{1}{x_{n}}\cdot \limsup x_{n}\geq 1.$$

I think the the simplest way to do that is to show that $$\liminf \frac{1}{x_{n}}=\frac{1}{\limsup x_{n}}.$$

I started to write some things, but nothing leaded me to the solution.

I'd like your help with this.

Thank you

Answer 1

Lemma
Suppose that $A\subset\mathbb{R}$ such that there exists $a,b\in\mathbb{R}^+$ so that $A$ is bounded below by $a$ and above by $b$. Let $\frac{1}{A}:=\left\{\frac{1}{x}:\ x\in A\right\}$. Then $\inf\frac{1}{A}=\frac{1}{\sup A}$.

From this it follows that $$\inf\left\{ \frac{1}{x_m}: m\geq n \right\}=\frac{1}{\sup\left\{ x_m:\ m\geq n \right\}}$$ and what you want to show follows by taking limits as $m\rightarrow\infty$. Also, note that it is very important that the $x_n$ are all positive and bounded away from both $0$ and infinity.

Note: It may be useful to recall the definition of $\limsup$: $$ \limsup_{n\to\infty}x_n := \lim_{n\to\infty}\Big(\sup_{m\geq n}x_m\Big)$$ (see Wikipedia)