$\limsup \sqrt[n]{a_n} \leq \limsup a_n$ if $(a_n)$ is non-negative sequence?

Question

Let $(a_n)$ be a sequence with $0 \leq a_n \leq 1$. Is it possible to show, that $\limsup \sqrt[n]{a_n} \leq \limsup a_n$?

Conversely, if $(a_n)$ is a sequence with $0 \leq a_n$ and $\limsup \sqrt[n]{a_n} \leq 1$. Is it possible to show, that there exists a $N$ so that $a_n \leq 1$ for all $n>N$.

Answer 1

$a_n=1/n$ and $a_n=n$ provide counterexamples to the two assertions. For that matter, so do $a_n=1/2$ and $a_n=2$.

Answer 2

No: take $a_n = 1/2^n$; then $\lim \sqrt[n]{a_n} = 1/2 > 0 = \lim a_n$.

No: take $a_n = (1+\tfrac1n)^n$; then $\lim \sqrt[n]{a_n} = 1$ but all $a_n>1$.