Simplifiying (or getting the exponent thing outside) the expression $\{({\rm complicated\, stuff})^{1/n}\}$ So i have like a very complicated expression involving many variables, and i want to simplify it, the only problem comes from the fact that this complicated expression is itself inside a fractional part $\{...\}$ which we define to be $$\{x\}={\rm fractional\,part\,of\,}x=x-{\rm IntegerPart}(x)$$
in particular this complicated expression takes the form $$\{({\rm complicated\, stuff})^{1/n}\}=({\rm complicated\, stuff})^{1/n}-{\rm IntegerPart}(({\rm complicated\, stuff})^{1/n})$$ I tried raising to the $n$ power and using newton's binomial theorem but we don't have things that get simplified
if anyone knows a way to write that in an alternative way (like getting rid of the sqrt) that would help very much!

TLDR; How to simplify $\{({\rm complicated\, stuff})^{1/n}\}$, or how to write $\{({\rm complicated\, stuff})^{1/n}\}=f({\rm complicated\, stuff})^{1/n}$

if anything is unclear feel free to ask!

the exact expression is $$\left\{\dfrac{1}{\left(1-\dfrac{p}{q}\right)^{1/n}}\right\}$$ p,q, integers
 A: I assume that
$0 < p < q$.
$\begin{array}\\
\dfrac{1}{\left(1-\dfrac{p}{q}\right)^{1/n}}
&=\dfrac{1}{\left(\dfrac{q-p}{q}\right)^{1/n}}\\
&=\left(\dfrac{q}{q-p}\right)^{1/n}\\
&=\left(\dfrac{q-p+p}{q-p}\right)^{1/n}\\
&=\left(1+\dfrac{p}{q-p}\right)^{1/n}\\
\end{array}
$
Since
$(1+x)^n
\ge 1+nx
$,
$(1+\frac{x}{n})^n
\ge 1+x
$
so that
$(1+x)^{1/n}
\le
1+\frac{x}{n}
$.
Therefore
$\left(1+\dfrac{p}{q-p}\right)^{1/n}
\le 1+\dfrac{p}{n(q-p)}
$
or
$\dfrac{1}{\left(1-\dfrac{p}{q}\right)^{1/n}}
\le 1+\dfrac{p}{n(q-p)}
= 1+\dfrac{p/q}{n(1-p/q)}
= 1+\dfrac{1}{n(q/p-1)}
$.
Does this help?
(added in response to a comment)
If
$\dfrac{1}{n(q/p-1)} < 1$,
then
$\left\lfloor \dfrac{1}{\left(1-\dfrac{p}{q}\right)^{1/n}} \right\rfloor
= 1$.
This happens if
$n
> \dfrac{1}{(q/p-1)}
= \dfrac{p}{q-p}
$.
Also,
by the generalized binomial theorem
if $n$ is large,
if $r = p/q$,
then
$(1-r)^{-1/n}
=1+\frac{r}{n}+\frac{r^2(1+n)}{2n^2}+...
$.
This doesn't behave well,
since the coefficient
of $r^m$
is about
$\frac{1}{m!n}
$.
It might be better to look at
$(1+s)^{1/n}
$
where
$s = \frac{p}{q-p}$.
This is
$(1+s)^{1/n}
=1+\frac{s}{n}-\frac{s^2(n-1)}{2n^2}+...
$.
Since this is an enveloping series,
$1+\frac{s}{n}-\frac{s^2(n-1)}{2n^2}
< (1+s)^{1/n}
< 1+\frac{s}{n}
$.
