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

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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

1 Answer 1

up vote 1 down vote accepted
+100

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

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Great response. Thank you for the help and insight. –  The Substitute Jan 24 '13 at 21:31

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