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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?

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  • $\begingroup$ As an example, would $\{5.4\} = 5.4 - [5.4] = 5.4 - 5 = 0.4$? $\endgroup$ – Andrew Shedlock Aug 21 at 17:07
  • $\begingroup$ Yes that is right $\endgroup$ – Keith Aug 21 at 17:08
  • $\begingroup$ It might help to represent the initial value explicitly, like: $$m+\frac{p}{q}$$ where all the numbers are natural and $p<q$ $\endgroup$ – Yuriy S Aug 21 at 17:10
  • $\begingroup$ I tried but without further information on $p$ and $q$ the calculations do not seem to proceed well... $\endgroup$ – Keith Aug 21 at 17:10
  • $\begingroup$ I just want to add that this does not converge to $round(a_0)$, can consider $a_0 = \frac{3}{4}$ $\endgroup$ – Andrew Shedlock Aug 21 at 17:38
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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.

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    $\begingroup$ We can say a bit more-which integer it converges to. You are almost there. $\endgroup$ – Ross Millikan Aug 21 at 17:44
  • $\begingroup$ 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? $\endgroup$ – Keith Aug 21 at 18:06
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    $\begingroup$ 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? $\endgroup$ – Keith Aug 21 at 18:20
  • $\begingroup$ I can't get it. To what you add bs?? $\endgroup$ – Keith Aug 21 at 18:38
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    $\begingroup$ 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 $11$. $\endgroup$ – automaticallyGenerated Aug 21 at 21:36
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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.

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

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