Let $ (c\_n)\_{n\geq 0} $ be a sequence of positive reals such that $ \dfrac{1}{m}\sum\_{k=0}^{m-1}c\_{k}\sim\prod\_{k=0}^{m-1} c\_{k}$ as $ m $ tends to infinity.
Call such a sequence a "corridor sequence" (as intuitively each of the terms should be "close to $ 1 $).


Which upper bound can be given for the quantity $ \sup\_{k\leq m}\{c\_{k}\}-\inf\_{k\leq m}\{c\_{k}\} $ provided the set of values of the corridor sequence is dense in some interval containing $ 1 $ ? 



Upper and lower bounds of sequences whose product of terms is asymptotically equal to their arithmetic mean
$\newcommand{\de}{\delta}
\newcommand{\De}{\Delta}
\newcommand{\ep}{\epsilon}
\newcommand{\ga}{\gamma}
\newcommand{\Ga}{\Gamma}
\newcommand{\la}{\lambda}
\newcommand{\Si}{\Sigma}
\newcommand{\thh}{\theta}
\newcommand{\R}{\mathbb{R}}
\newcommand{\F}{\mathcal{F}}
\newcommand{\E}{\operatorname{\mathsf E}} 
\newcommand{\PP}{\operatorname{\mathsf P}}
\newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$ 


Let us also use a comment by the OP, stating that the interval (say $I$) in which the $c\_k$'s are dense is of nonzero length and only finitely many $c\_k$'s are outside of $I$. 


Let us then show that such a sequence of $c\_k$'s cannot exist (and so, any bound on $\sup\_{k\leq m}\{c\_{k}\}-\inf\_{k\leq m}\{c\_{k}\} $ will be true). (The main idea here is that the consecutive arithmetic means may only vary very slowly, whereas the consecutive products may vary very fast.) 


Indeed, suppose the contrary, that there is a sequence $(c\_n)\_{n\ge0}$ of positive reals such that $$a\_m:=\dfrac{1}{m}\sum\_{k=0}^{m-1}c\_{k}\sim\prod\_{k=0}^{m-1} c\_{k}=:b\_m$$ 
as $m\to\infty$, the $c\_k$'s are dense in an interval $I$ of nonzero length containing $1$, and only finitely many $c\_k$'s are outside of $I$. Then there is some $\de\in(-1,1)\setminus\{0\}$ such that $1+\de\in I$. Therefore and because the $c\_k$'s are dense in $I$, 
\begin{equation\*}
 c\_m=(1+\de)(1+o(1)) \tag{1}
\end{equation\*}
infinitely often (i.o.), that is, $c\_{m\_j}=(1+\de)(1+o(1))$ for some increasing sequence $(m\_j)$ of natural numbers as $j\to\infty$. So, 
\begin{equation\*}
 b\_{m+1}=b\_mc\_m=b\_m(1+\de)(1+o(1))=a\_m(1+\de)(1+o(1)) \tag{2}
\end{equation\*}
i.o. On the other hand,  

\begin{equation\*}
 a\_{m+1}=\frac{m}{m+1}a\_m+\frac{c\_m}{m+1}=a\_m(1+o(1))+\frac{c\_m}{m+1}. \tag{3}
\end{equation\*}
Since $b\_{m+1}=a\_{m+1}(1+o(1))$, (2) and (3) imply that i.o.
\begin{equation\*}
 a\_m(1+\de)(1+o(1))=a\_m(1+o(1))+\frac{c\_m}{m+1}, 
\end{equation\*}
whence $a\_m[(1+\de)(1+o(1))-(1+o(1))]=\frac{c\_m}{m+1}\sim\frac{1+\de}m$ and 
\begin{equation\*}
 a\_m\sim\frac{1+\de}\de\,\frac1m. \tag{4}
\end{equation\*}
Since $a\_m>0$, it follows that necessarily $\de>0$, that is, $I\subseteq[1,\infty)$. 


Since only finitely many $c\_k$'s are outside of $I$, we see that the positive numbers $b\_m$ are nondecreasing in all large enough $m$, and so, $b\_m$ goes to some $b\in(0,\infty]$ as $m\to\infty$. But this contradicts (4), since $a\_m\sim b\_m$. 


