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Iosif Pinelis
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$\newcommand{\de}{\delta} \newcommand{\De}{\Delta} \newcommand{\ep}{\epsilon} \newcommand{\ga}{\gamma} \newcommand{\Ga}{\Gamma} \newcommand{\la}{\lambda} \newcommand{\Si}{\Sigma} \newcommand{\thh}{\theta} \newcommand{\R}{\mathbb{R}} \newcommand{\F}{\mathcal{F}} \newcommand{\E}{\operatorname{\mathsf E}} \newcommand{\PP}{\operatorname{\mathsf P}} \newcommand{\ii}[1]{\operatorname{\mathsf I}\{#1\}}$

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).

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 $a_m\sim b_m$.

Iosif Pinelis
  • 127.8k
  • 8
  • 107
  • 229