6
$\begingroup$

Consider a sequence of positive real numbers $(a_n)$. Define $[a_1]=\frac{1}{a_1}$ and recursively inductively $[a_1,\cdots,a_n]=\frac{1}{a_1+[a_2,\cdots,a_n]}$. Suppose $a_k\geq 2$ for all $k$. How to show that $\lim_{n\to\infty}[a_1,\cdots,a_n]$ exists?

I was trying to show that the sequence is monotone, which is not true. A special related case is done here, which is not very helpful to have a generalization.


[Added to answer the confusion in comments.] The definition above should be understood properly as follows. For any positive real number $a$, define $[a]:=\frac{1}{a}$. Now, given any two positive real numbers $a_1,a_2$, one can define $[a_1,a_2]:=\frac{1}{1+[a_2]}$. One can thus keep going on in this fashion to define $[a_1,a_2,\cdots,a_n]$. To write down a few terms explicitly, $$ [a_1,a_2]=\frac{1}{a_1+\frac{1}{a_2}},\ [a_1,a_2,a_3]=\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3}}},\ [a_1,a_2,a_3,a_4]=\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{a_4}}}},\cdots $$

$\endgroup$
9
  • $\begingroup$ Your sequence is not well defined $\endgroup$
    – matboy
    Jun 28, 2017 at 13:47
  • 1
    $\begingroup$ @matboy: Why not? $\endgroup$
    – user9464
    Jun 28, 2017 at 13:48
  • $\begingroup$ Maybe because you ought to use $[a_1,\dots a_{n-1}]$ instead of $[a_2,\dots a_n]$ when you define your sequence. $\endgroup$ Jun 28, 2017 at 13:49
  • 1
    $\begingroup$ @uniquesolution: it is defined as it is written. What goes wrong logically with the definition? $\endgroup$
    – user9464
    Jun 28, 2017 at 13:50
  • 1
    $\begingroup$ What is $[a_1,a_2]$? According to your definition, it is $\frac{1}{a_1+[a_2]}$. So what is $[a_2]$? And then, what is $[a_1,a_2,a_3]$? you say it is $\frac{1}{a_1+[a_2,a_3]}$. But where do we get $[a_2,a_3]$ from ? Therefore, your sequence is not well defined as it is written. In fact, it is not recursive at all. $\endgroup$ Jun 28, 2017 at 13:54

2 Answers 2

3
$\begingroup$

You only need $a_i \ge 1$.

Define $$\begin{bmatrix} p_{-1} & p_{0} \\ q_{-1} & q_{0} \end{bmatrix} = \begin{bmatrix} 1 & 0 \\ 0 & 1 \end{bmatrix}. $$ Recursively, set $$p_k = a_kp_{k-1} + p_{k - 2}, q_k = a_kq_{k-1} + q_{k - 2} \tag{1} $$ for $k \ge 1$. We will show that

$$ [a_1,\dots,a_n,a_{n+1}] = \frac{a_{n+1}p_n + p_{n - 1}}{a_{n+1}q_n + q_{n - 1}}. $$

This holds by induction:

\begin{align} [a_1,\dots,a_n,a_{n+1}] &= \left[ a_1,\dots,a_n + \frac{1}{a_{n+1}} \right] \\ &= \frac{\left(a_n + \frac{1}{a_{n+1}} \right)p_{n-1}+p_{n-2}}{\left(a_n + \frac{1}{a_{n+1}} \right)p_{n-1}+p_{n-2}} \\ &= \frac{(a_np_{n-1}+p_{n-2)} + \frac{1}{a_{n+1}}p_{n - 1}}{(a_nq_{n-1}+q_{n-2}) + \frac{1}{a_{n+1}}q_{n - 1}} \\ &= \frac{a_{n+1}p_n + p_{n - 1}}{a_{n+1}q_n + q_{n - 1}} \end{align} and the base case is easy to verify.

Thus we get the matrix equation $$ \begin{bmatrix} p_n & p_{n-1} \\ q_n & q_{n - 1} \end{bmatrix} \begin{bmatrix} a_{n+1} & 1 \\ 1 & 0 \end{bmatrix} = \begin{bmatrix} p_{n+1} & p_{n} \\ q_{n+1} & q_{n} \end{bmatrix} $$ and by induction $$ \begin{bmatrix} p_{n+1} & p_{n} \\ q_{n+1} & q_{n} \end{bmatrix} = \begin{bmatrix} a_{n+1} & 1 \\ 1 & 0 \end{bmatrix} \begin{bmatrix} a_{n} & 1 \\ 1 & 0 \end{bmatrix} \cdots \begin{bmatrix} a_{1} & 1 \\ 1 & 0 \end{bmatrix}\begin{bmatrix} 0 & 1 \\ 1 & 0 \end{bmatrix} $$ taking determinants we have $$ p_{n+1}q_n - p_nq_{n+1} = (-1)^{n}. $$ Hence $$ \left\lvert \frac{p_{n+1}}{q_{n + 1}} - \frac{p_n}{q_n} \right\rvert = \frac{|p_{n+1}q_n - p_nq_{n+1}|}{q_{n+1}q_n} = \frac{1}{q_{n+1}q_n}. $$

From $(1)$ and the assumption that $a_k \ge 1$ we have $q_n \ge n$. Thus the sequence $p_n/q_n$ is Cauchy and converges to some limit.

Reference

Automatic Sequences by J-P. Allouche and J. Shallit, Cambridge University Press (2003), pages 44-46.

$\endgroup$
2
  • $\begingroup$ Thank you for your answer and the reference. It seems that you have generalized the result in the quoted book, in which the $a_i$'s are required to be integers. $\endgroup$
    – user9464
    Jun 28, 2017 at 19:39
  • $\begingroup$ @Jack What you need is some condition on $a_k$ to allow you to conclude from the recursion $q_k = a_kq_{k - 1} + q_{k - 2}$ that $q_{n + 1}q_n$ is sufficiently large so that you end up with a Cauchy sequence. If $a_k \ge 1$ then $q_k \ge q_{k - 1} + q_{k - 2}$ and you can get some sort of weak bound from there. I believe the condition that $a_k \in \mathbf{Z}$ doesn't play a role in the proof and is only mentioned because the book is interested only in continued fractions over the integers. $\endgroup$ Jun 28, 2017 at 20:00
0
$\begingroup$

In Khinchin's book "Continued Fractions" it is shown that $$ \lim_{n\to\infty} [a_0;,a_1, \ldots, a_n] $$ exists if and only if $\sum_{n=1}^{\infty} a_n = \infty$, where $(a_n)_n$ is a sequence of positive numbers (Theorem 10). This solves your problem, since your sequence of positive numbers has a uniform lower positive bound.

$\endgroup$
1
  • $\begingroup$ Thanks for the reference! It seems that the proof of Theorem 10 is highly "non-trivial" while the statement is very simple. $\endgroup$
    – user9464
    Jun 28, 2017 at 19:36

You must log in to answer this question.