# Upper bound of some complicated function

Related to my finance research, I am trying to find a non-trivial upper bound of $$\begin{equation*} \left( \frac{ \left(1 - r^{\frac{1}{K-1}}\right)K}{ \left(1 - r^{\frac{K}{K-1}} \right) } -1\right)\cdot \frac{r}{1-r}, \end{equation*}$$ when we consider all $$r \in (0,1)$$ and $$K$$ being any integer weakly greater than $$2$$. (A numerical exercise suggests that it can be bounded from above by $$0.5$$, and the bound is achieved as $$r$$ goes to $$1$$.)

What I know is that it is non-negative and bounded from above by $$1$$, because it comes from the share of one trader's profit relative to the sum of profits of two or more traders, all of which are positive.

• What does "weakly greater than $2$" mean, please? Jun 24, 2020 at 14:10
• Thanks for your comment. I meant that K can be any integer that is equal to or greater than 2, when we consider an upper bound. Jun 24, 2020 at 14:16

Have you try proving that for any $$K>0$$:
$$\lim_{r\to1 ^{-}} \left( \frac{ \left(1 - r^{\frac{1}{K-1}}\right)K}{ \left(1 - r^{\frac{K}{K-1}} \right) } -1\right)\cdot \frac{r}{1-r} =0.5?$$
That should give you the $$1/2$$ bound that you seem to be obtaining. Should be easy since for your parameter values this function is continous and monotonously increasing in $$r$$.
• I agree with your point. I'm trying to prove that it is increasing in $r$, but my attempt has not been successful. Jun 24, 2020 at 16:09
$$\begin{equation*} y=\left( \frac{ \left(1 - r^{\frac{1}{k-1}}\right)k}{ \left(1 - r^{\frac{k}{k-1}} \right) } -1\right)\, \frac{r}{1-r} \end{equation*}$$ The Taylor expansion built around $$r=1$$ is $$y=\frac{1}{2}-\frac{2 k-1 }{12 (k-1)}(1-r)-\frac{2 k-1 }{24 (k-1)}(1-r)^2+O\left((1-r)^3\right)$$ and you have $$y <\frac{1}{2}-\frac{2 k-1 }{12 (k-1)}(1-r)$$
In fact, it seems (at least up to degree $$5$$) that the expansion is an upper bound.