I have proved this theorem, can somebody check? So
$\textbf{Theorem:}$ Let $(x_n)$ be a bounded sequence, then $(x_n)$ convergegs if and only if $\limsup(x_n)=\liminf(x_n)$.
$\textbf{Proof:}$ "$\Rightarrow$" Let $(x_n)$ converges to $x$. So $\exists \ N_1 \ \in \ \mathbb N$ such that $|x_n-x|< \epsilon \ \forall \ n\geq N_1$. So we have that $x-\epsilon<x_n<x+\epsilon$. Hence $x-\epsilon<x_n$ for infinite number $n \ \in \ \mathbb N$, then $x=\limsup(x_n)$ (by definition of $\limsup$). On the other hand we have $x_n<x+\epsilon$ for infinite number $n \ \in \ \mathbb N$, then $x=\liminf(x_n)$ (by definition of $\liminf$).
Another direction was much easier, therefore I need help only with 1 direction. Thank you!
Best Answer
You have the basic idea, but I find it easier to go from these definitions: \begin{align} \liminf_{n\to\infty} x_n &= \lim_{n\to\infty} \inf_{m\geqslant n} x_m\\ \limsup_{n\to\infty} x_n &= \lim_{n\to\infty} \sup_{m\geqslant n} x_m. \end{align} So if $\lim_{n\to\infty} x_n=x$, then for all $\varepsilon>0$ there exists $N$ such that $$x-\varepsilon < \inf_{m\geqslant N}x_m < \sup_{m\geqslant N}x_m<x+\varepsilon,$$ and hence these quantities are equal.