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, $$ [a_1,a_2]=\frac{1}{a_1+\frac{1}{a_2}},\ [a_1,a_2,a_3]=\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3}}},\ [a_1,a_2,a_3,a_4]=\frac{1}{a_1+\frac{1}{a_2+\frac{1}{a_3+\frac{1}{a_4}}}},\cdots $$

$\endgroup$
  • $\begingroup$ Your sequence is not well defined $\endgroup$ – matboy Jun 28 '17 at 13:47
  • 1
    $\begingroup$ @matboy: Why not? $\endgroup$ – Jack Jun 28 '17 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. $\endgroup$ – uniquesolution Jun 28 '17 at 13:49
  • 1
    $\begingroup$ @uniquesolution: it is defined as it is written. What goes wrong logically with the definition? $\endgroup$ – Jack Jun 28 '17 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. $\endgroup$ – uniquesolution Jun 28 '17 at 13:54
4
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – Jack Jun 28 '17 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. $\endgroup$ – Trevor Gunn Jun 28 '17 at 20:00
1
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ Thanks for the reference! It seems that the proof of Theorem 10 is highly "non-trivial" while the statement is very simple. $\endgroup$ – Jack Jun 28 '17 at 19:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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