$\theta$ 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 $a_{k+1}>N$
A Question on Using a Half-Divergent Sequence.
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0what is $n_{k_1}$ in your problem? – 2012-12-09
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0The first term of the sequence $(n_k)$ of denominators of the sequence of principal convergents of some irrational number in $(0,1).$ The $(n_{k_i})$ is a subsequence. – 2012-12-09
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0I've edited the problem. I don't mind losing points, but explaining the downvote would be helpful. – 2012-12-09
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0sorry. slip of stupid mouse.meant to upvote.i can't upvote until the question is edited again.will do soon – 2012-12-09
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0Sorry for the many edits I have made. I tried making the problem more accessible, but I realized that I was completely altering the problem by doing so. – 2012-12-10
1 Answers
For any $k_i$ such that $0\le k_1
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 $\delta>0$, $$\lim_i \frac{\log a_{k_i+1} }{n_{k_{i+1}}-n_{k_i}}=0, \qquad {\rm for\ all\ } k_1
$$\limsup_i \frac{\log a_{k_i+1} }{n_{k_{i+1}}-n_{k_i}}>0.$$
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0Great response. Thank you for the help and insight. – 2013-01-24