Help proving a limit exists using limsup and liminf.

Question

Let $X=(x_n)_{n\in N}$ be a bounded sequence of real numbers with $\liminf_{n \to \infty}x_n=\limsup_{n \to \infty}x_n$. Prove that $X$ is convergent.

I'm not sure where to start. Any suggestions?

Answer 1

Let $L^+= \limsup x_n$ and $L^-=\liminf x_n$. Then both are real numbers because the sequence is bounded by hypothesis.

Given $\varepsilon>0$, so there is some $N$ such that $x_n<L^++\varepsilon$ and $L^--\varepsilon< x_n$ for all $n\ge N$. Thus $L^--\varepsilon<x_n<L^++\varepsilon$. Since $L^-=L^+$, we therefore have

$$L-\varepsilon<x_n<L+\varepsilon$$

i.e., $|x_n-L|< \varepsilon$, where $L=L^+=L^-$.