Does convergence of arithmetic mean imply convergence of geometric mean? Let $x_n$ be a sequence of positive real numbers, which is not convergent. ($x_n$ does not converge to a finite number, nor to infinity). Define
$$ A_n = \frac{x_1 + x_2 + \cdots + x_n}{n} \quad \text{ and }\quad G_n = \sqrt[n]{x_1x_2...x_n}.$$

Does the convergence of $A_n$ imply the convergence of $G_n$ or vice versa?

I know that if $x_n \to L \in \mathbb{R}\cup\{\infty\}$, then both $A_n,G_n$ converge to $L$. But here I assume $x_n$ is not convergent.
I also wonder if adding a boundedness assumption on $x_n$ changes anything.

Here's an example to show that the convergence of $A_n$ does not imply the convergence of $G_n$.

Let $\mathbb{Z}_0$ denote the set of nonnegative integers, and let
$S=\{2^k{\,\mid\,}k\in\mathbb{Z}_0\}$.

Let the sequence $x_1,x_2,x_3,...$ be defined by
$$
x_n=
\begin{cases}
{\large{\frac{1}{2^n}}}&\text{if}\;n\in S\\[4pt]
1&\text{otherwise}
\end{cases}
$$
Then the sequence $(x_n)$ is a bounded, non-convergent sequence of positive real numbers.

For $A_n$ we have
$$
\frac{n-\bigl(\log_2(n)+1\bigr)}{n} < A_n < 1
$$
hence ${\displaystyle{\lim_{n\to\infty}}}A_n=1$.

For $G_n$ we have
$$
G_n
=
\begin{cases}
{\large{\frac{1}{2}}}&\text{if}\;n+1\in S\\[4pt]
{\Large{\frac{{\Large{1}}}{2^{\left(2-{\large{\frac{1}{n}}}\right)}}}}&\text{if}\;n\in S\\
\end{cases}
$$
hence the sequence $(G_n)$ has limit points ${\large{\frac{1}{2}}}$ and ${\large{\frac{1}{4}}}$, so $(G_n)$ is non-convergent.
