Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

$\theta$ 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 $a_{k+1}>N$

share|cite|improve this question
what is $n_{k_1}$ in your problem? – dexter04 Dec 9 '12 at 9:16
The 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. – The Substitute Dec 9 '12 at 9:20
I've edited the problem. I don't mind losing points, but explaining the downvote would be helpful. – The Substitute Dec 9 '12 at 12:25
sorry. slip of stupid mouse.meant to upvote.i can't upvote until the question is edited again.will do soon – dexter04 Dec 9 '12 at 12:27
Sorry 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. – The Substitute Dec 10 '12 at 2:20
up vote 1 down vote accepted

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:

  1. 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.

  2. 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 $$\limsup_i \frac{\log a_{k_i+1} }{n_{k_{i+1}}-n_{k_i}}>0.$$

share|cite|improve this answer
Great response. Thank you for the help and insight. – The Substitute Jan 24 '13 at 21:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.