A Question on Using a Half-Divergent Sequence. $\theta$ 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 $n_k^\delta<a_k<2n_k^\delta$.
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 $a_{k+1}>N$
 A: For any $k_i$ such that $0\le k_1<k_2<\dots$ and all $i\ge 1$,
$$
n_{k_{i+1}}-n_{k_i}\ge n_{k_{i+1}}-n_{k_{i+1}-1}\ge (a_{k_{i+1}}-1) n_{k_{i+1}-1},
$$
and by an easy induction,
$$
n_k\ge a_1\dots a_k \qquad {\rm for\ all\ }k\ge 1,
$$
so if $i\ge 2$,
$$
n_{k_{i+1}}-n_{k_i}\ge a_1\dots a_{k_{i+1}-1} (a_{k_{i+1}}-1). \qquad (*)
$$
In any continued fraction, the sequence $(n_k)$ must grow at least at an exponential rate.  Combined with the hypothesis $a_k>n_k^\delta$, this implies that the sequence $(a_k)$ grows at least at an exponential rate.  Therefore, there is some $A>1$ and $k_0$ such that $a_k\ge A^k\ge 2$ for all $k\ge k_0$.  Now, using this with (*), if $i\ge 2$ is sufficiently large so that $k_i+1\ge k_0$,
$$
\frac{\log a_{k_i+1} }{n_{k_{i+1}}-n_{k_i}}\le 
\frac{\log a_{k_i+1}}{a_1\dots a_{k_{i+1}-1} (a_{k_{i+1}}-1)}
\le \frac{\log a_{k_i+1}}{a_{k_i+1}-1}.\qquad (**)
$$
However, since $a_k$ grows exponentially with $k$, the right-hand side of (**)
decreases exponentially with $k_i+1$.  Therefore, the left-hand side of (**) has limit $0$.
Summarizing the above:


*

*Given the hypothesis $a_k>n_k^\delta$ for some $\delta>0$, $a_k$ increases at least exponentially with $k$, so $\lim_k a_k=\infty$, and $(a_k)$ is trivially half-divergent.

*Given the hypothesis $a_k>n_k^\delta$ for some $\delta>0$, $$\lim_i \frac{\log a_{k_i+1} }{n_{k_{i+1}}-n_{k_i}}=0, \qquad
{\rm for\  all\ } k_1<k_2<\dots,$$
so there is no way to choose $(k_i)$ so that 
$$\limsup_i \frac{\log a_{k_i+1} }{n_{k_{i+1}}-n_{k_i}}>0.$$
