# limit of sequence defined by rationals and fractional parts

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?

• As an example, would $\{5.4\} = 5.4 - [5.4] = 5.4 - 5 = 0.4$? – Andrew Shedlock Aug 21 at 17:07
• Yes that is right – Keith Aug 21 at 17:08
• 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 at 17:10
• I tried but without further information on $p$ and $q$ the calculations do not seem to proceed well... – Keith Aug 21 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 at 17:38

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.

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

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.

• 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 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 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 at 17:25