Understanding limit superior Given that $x_n$ is a real sequence, how does one show that  $$\overline{\lim_{n\to \infty}} x_n =\lim_{n\to \infty}(\sup_{k\geq n}x_k).$$
I know the above is the definition for limit superior so I'm a bit confused.
 A: The definition of limit superior is as follows:
Let $\left(x_n\right)$ be a bounded sequence of real numbers and define a sequence $(y_n)$,
 \begin{equation}y_n=\sup_{k\geq n}x_k=\sup \left\{ x_k: k\ge n \right\}=\sup \left\{ x_n,{x_{n+1}},... \right\}\end{equation}
By the Least Upper Bound property $(y_n)$ is well defined. Since $\forall n\in \mathbb{N}\ x_n\le y_n$, $(y_n)$ is bounded below (by say $x_1$). In addition, $(y_n)$ is decreasing (why?*) and thus by the Monotone Convergence Theorem, 
 \begin{equation}y_n\to \inf_ny_n\end{equation}
We define 
\begin{equation}\lim\sup x_n:=\inf_ny_n=\displaystyle\lim_{n\to \infty}y_n=\lim_{n\to \infty}(\sup_{k\geq n}x_k)\end{equation}
*Hint: If $A\subset B\subset \mathbb{R}$ and $A,B$ are bounded then $\sup A\le \sup B$
A: Consider $y_n=\sup\limits_{k\ge n} x_k$. Show that the $y_n$ form a nonincreasing sequence (using the definition of supremum). Hence, $$\lim\limits_{n\to\infty}\left(\sup\limits_{k\geq n}x_k\right)=\lim\limits_{n\to\infty}y_n=\inf\limits_n y_n=\overline{\lim\limits_{n\to\infty}}x_n.$$
