Properties of limsup $\limsup \sqrt{a_n} \le \sqrt{\limsup a_n}$

Question

Define $\limsup a_n = \lim_{n \to \infty} \sup \{a_n, a_{n+1},\ldots\}$.

Let $(a_n)$ be a positive bounded sequence. Prove or disprove that $\limsup_{n \to \infty} \sqrt{a_n} \le \sqrt{\limsup_{n \to \infty} a_n}$.

Answer 1

As an alternative (yet equivalent) definition, an upper limit of a sequence $\{x_n\}$ is the sup of the set of its subsequential limits (which is a closed set as illustrated in the same link above).

By this definition, suppose $\limsup a_n=a$, then $\sup A= a$ where $A$ denotes the set of all the subsequential limits of $\{a_n\}$. This also tells that $a\in A$. Thus, we can pick a subsequence $\{a_{n,k}\}$ such that $a_{n,k}\to a$ as $k\to\infty$, but we cannot pick any subsequence that converges to a number greater than $a$.

Use these facts to show equality holds in your example.