Since $a_n\rightarrow \infty$ we may disregard the first few elements and assume $a_1>0$. Now suppose that $\sum_1^\infty (1-{{a_n}\over {a_{n+1}}})<\infty$. All terms are contained in $[0, 1)$. Let $S$ be the set of $n$ for which $a_n\not=a_{n+1}$, then $S$ is infinite and $\sum_S (1-{{a_n}\over {a_{n+1}}})<\infty$. Hence $\Pi_S {{a_n}\over {a_{n+1}}}>0$ by a well-known convergence criterion for infinite products. Hence also $\Pi_1^\infty {{a_n}\over {a_{n+1}}}>0$ since we only add factors 1. This is a contradiction since $\Pi_1^n {{a_k}\over {a_{k+1}}}=a_1/a_{n+1}\rightarrow 0$.

Suppose the contrary that for some $M\in(0,\infty)$, \[ b_n = \sum_{1\le k\le n-1}\left(1-\frac{a_i}{a_{i+1}}\right) <M,\quad \forall n. \]Let $c_i = 1-a_i/a_{i+1}$. It is evident that $c_i\to 0$, so there is an $N$ such that $c_i<1/2$ for $i\ge N$. Now, check that $\ln(1-x) \ge -2x$ for $x\in[0,1/2]$. So, for any $L>N$, \[ \sum_{N\le k\le L-1} \ln(1-c_k)\ge -2\sum_{N+1\le k\le L-1}c_k \ge -2M\Rightarrow a_L\le a_N e^{2M}. \]Since $L$ is arbitrary, this contradicts with the fact $(a_n)_{n\ge 1}$ is bounded.