Notice that $$b_k=\sup_{n\in \mathbb{Z}}\left\{\left \lceil \frac{a_n+a_{n-1}+\dots+a_{n-k+1}}{a_n} \right \rceil\right\}\;(k=1,2,\dots),$$so $b_k\le k$ and also $b_k\le b_{k+1}$. Suppose $b_1,b_2,b_3,\dots$ does not coincide with $1,2,3,...$, then there exists a smallest $m$ such that $b_{m+1}<m+1$. It's clear that $b_1=1, b_2=2$, so $m\ge 2$ and from $b_k$ is non-decreasing we know that $b_{m+1}=m$. To prove that $\{b_k\}$ is eventually constant, it suffices to prove that ${b_k}$ is bounded. Since $b_{m+1}=m$, for every $n$ the inequality $$ma_n\ge a_n+a_{n-1}+\dots+a_{n-m}>(m+1)a_{n-m},$$holds. Let $c=\frac{m}{m+1}<1$, then from induction we obtain $$a_{n-tm}<c^ta_m$$for any $n\in \mathbb{Z}$ and $t\in \mathbb{Z_+}$. Summing up gives $$\sum_{\substack{n'\le n\\n'\equiv n \;(\text{mod}\; m)}}a_{n'}<\frac{1}{1-c}a_n.$$Therefore \begin{align*}\sum_{n'\le n}a_{n'} &=\sum_{i=0}^{m-1}\left (\sum_{\substack{n'\le n-i\\n'\equiv n-i \;(\text{mod}\; m)}}a_{n'} \right )\\ &<\sum_{i=0}^{m-1}\frac{1}{1-c}a_{n-i}.\\ &\le \frac{m}{1-c}a_n. \end{align*}This means $b_k<\frac{m}{1-c}$ for all $k$, and by $\{b_k\}$ is a non-decreasing integer sequence, $\{b_k\}$ is eventually constant.