Assume otherwise. For any $n$ there is $t=t_n$ such that \[ \frac{a_n}{a_{n+1}}+\frac{a_{n+1}}{a_{n+2}}+\cdots+\frac{a_{n+t-1}}{a_{n+t}} < t-\varepsilon, \]But from AM-GM inequality, $LHS \ge t \left( \frac{a_n}{a_{n+t}} \right)^{1/t}$, therefore \[ \frac{a_n}{a_{n+t}} < \left( 1-\frac{\varepsilon}{t} \right)^t <e^{-\varepsilon}. \]Analagously, there exists $t'$ such that $\frac{a_{n+t}}{a_{n+t+t'}} <e^{-\varepsilon}$, and so on. Multiplying, we get $\frac{a_n}{a_{\text{something}}} <e^{-N \varepsilon} < \frac{1}{M}$, which is absurd.

Assume for every $m$ there exists $f(m)$ such that $\sum\limits_{j=m}^{f(m)-1} \frac{a_j}{a_{j+1}} < f(m)-m-\epsilon$. The idea is to show for a fixed constant $n$, there doesn't exist a chain $m\to f(m) \to \cdots \to f^n(m)$. Then, $\sum\limits_{j=m}^{f^n(m)-1} \frac{a_j}{a_{j+1}} < f^n(m)-m-n\epsilon$. For simplicity, let $d=f^n(m)-m$. By AM-GM, $\frac{a_m}{a_{f^n(m)}} = \prod\limits_{j=m}^{f^n(m)-1} \frac{a_j}{a_{j+1}} < (\frac 1d \sum\limits_{j=m}^{f^n(m)-1} \frac{a_j}{a_{j+1}})^d < (1-\frac{n\epsilon}{d})^d$. As $d\to \infty$, the last expression tends to $e^{-n\epsilon}$ and is smaller than $2^{-n\epsilon}$. Therefore, picking $n$ such that $2^{-n\epsilon} < \frac 1M$ gives a contradiction.