limit of sequence defined by rationals and fractional parts

Question

Let $a_0$ be a positive rational number. And for natural numbers $n$, define a sequence as $a_n=a_{n-1}/(1-\{a_{n-1}\})$ where $\{x\} = x-[x]$ , the fractional part of $x$. Then I have to show that the limit of this sequence exists and find the limit. However, the fractional part makes everything very tricky....Could anyone please help me?

It might help to represent the initial value explicitly, like: $$m+\frac{p}{q}$$ where all the numbers are natural and $p<q$ — Yuriy S, Aug 21 '19 at 17:10
I tried but without further information on $p$ and $q$ the calculations do not seem to proceed well... — Keith, Aug 21 '19 at 17:10
I just want to add that this does not converge to $round(a_0)$, can consider $a_0 = \frac{3}{4}$ — Andrew Shedlock, Aug 21 '19 at 17:38
Have you made a spreadsheet and tried a bunch of $a_0$s? Copy down makes it easy. Playing a bit can lead you to a hypothesis. — Ross Millikan, Aug 21 '19 at 17:42

Varun Vejalla · Accepted Answer · 2019-08-21T21:33:45.077

5

The only time $a_n = a_{n-1}$ is when $a_{n-1}$ is an integer. This is because $$a_n = \frac{a_{n-1}}{1-0} = a_{n-1}$$

Now we simply have to show that the denominator of $a_n$ is continuously decreasing when the denominator is not equal to $1$. Assume that $a_{n-1} = \frac{mb+p}{b}, 0 < p < b, \gcd(mb+p, b) = 1$, and all are natural numbers. Then we have that $$a_n = \frac{\frac{mb+p}{b}}{\frac{b-p}{b}} = \frac{mb+p}{b-p}$$ Since $b-p < b$, the denominator decreases. Thus, $a_n$ will converge to an integer.

edited Aug 21 '19 at 21:33

answered Aug 21 '19 at 17:39

Varun Vejalla

8,873

1

We can say a bit more-which integer it converges to. You are almost there. – Ross Millikan Aug 21 '19 at 17:44
Ok thanks to your help, I was able to understand that the denominator is forced to converge to 1. Thus the sequence becomes constant(an integer) at some point. However, I cannot grasp what that constant must be. Could you explain more please? – Keith Aug 21 '19 at 18:06
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I see. The final obstacle for me is that (because I am very not good at number theory) why $mb+p$ and $b-p$ being coprime is implied by $mb+p$ and $b$ being coprime? – Keith Aug 21 '19 at 18:20
I can't get it. To what you add bs?? – Keith Aug 21 '19 at 18:38
I think $mb+p$ and $b-p$ need not be coprime, but $mb+p$ and $b$ must be coprime. I think I can make my argument based on these facts. – Keith Aug 21 '19 at 21:32
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Sorry, but my earlier reasoning regarding what $a_n$ converges to was wrong. It doesn't necessarily have to converge to the numerator. Take $a_0 = \frac{55}{24}$ for example, which converges to $11$. – Varun Vejalla Aug 21 '19 at 21:36

Andrew Shedlock · Answer 2 · 2019-08-21T17:35:37.043

3

We can start by noticing that every for every real $x> 0$ that $0\leq {x} < 1$ so $$\frac{1}{1 - \{x\} }\geq 1$$ With equality only when $x$ is an integer. Hence if $a_n > 0$ then by induction we will have that $$ a_{n-1} < \frac{a_{n-1}}{1 - \{a_{n-1}\}} = a_n$$ When $a_{n-1}$ is not an integer and $a_{n-1} = a_n$ when $a_{n-1}$ is an integer. So if $a_0$ is an integer, then $a_n\rightarrow a_0$ and we need to consider now when $a_0$ is not an integer. Well since $a_0\not\in\mathbb{Z}$ then the sequence $\{a_n\}$ will end up being strictly increasing.

edited Aug 21 '19 at 17:35

answered Aug 21 '19 at 17:17

Andrew Shedlock

2,559

Yes I noticed that fact as well. I suspect that the sequence becomes an integer at some point. But I cannot prove it though. – Keith Aug 21 '19 at 17:18
If we start with an integer then it ends up being constant. So lets look at the numbers between two integers? – Andrew Shedlock Aug 21 '19 at 17:19
You need to escape the braces in the first line with backslashes as you have in the rest of the post – Ross Millikan Aug 21 '19 at 17:25

limit of sequence defined by rationals and fractional parts

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