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?
 A: 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.
A: 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.
