Upper bound of some complicated function

Question

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.

Thanks in advance for your help!

Answer 1

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 $r$.

Answer 2

$$\begin{equation*} y=\left( \frac{ \left(1 - r^{\frac{1}{k-1}}\right)k}{ \left(1 - r^{\frac{k}{k-1}} \right) } -1\right)\, \frac{r}{1-r} \end{equation*}$$ 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.