$\theta$ is an irrational in $[0,1]$ with continued fraction representation $[0;a_1,a_2,\dots]$, and the sequences $(a_k), (n_k)$ are related by the recurrence relation $n_{k+1}=a_{k+1}n_k+n_{k-1}, n_0=1, n_{-1}=0$. They are also related by the fact that there is a $\delta>0$ for which $n_k^\delta
Suppose $(a_k)$ is half-divergent (see $(*)$ below for my definition).
Suppose that for any $i$ I have $ s\ge \frac{\log_{a_{k_i+1}}}{n_{k_{i+1}}-n_{k_i}},$ where $n_{k_{i+1}}-n_{k_i}\to \infty$ and $(a_{k_i})$ is a subsequence of $(a_k)$
Hence $s\ge\limsup_{i\to \infty} \frac{\log(a_{k_i+1})}{n_{k_{i+1}}-n_{k_i}}$
Since $(a_k)$ is half-divergent, it diverges on any subsequence where it is unbounded. My question: Does this imply that for any $\varepsilon \le s$, I can choose a subsequence of $(a_k)$ for which $\varepsilon \le \limsup_{i\to \infty} \frac{\log(a_{k_i+1})}{ n_{k_{i+1}}- n_{k_i}}\le s?$ If so, is there an effective way to choose the subsequence?
$(*)$ A sequence $(a_k)$ is half-divergent if $\exists M\in \mathbb{R}$ such that $\forall N>M, \exists k_0$ such that $k>k_0$ implies that either $a_{k+1}\le M$ or $a_{k+1}>N$