Values of parameter $\epsilon \in (0,1)$ that make a rational function decreasing For $p \in (0,1]$, an integer $n \geq 2$ and $\epsilon \in (0,1)$, I want to show that  $$\frac{p (1- \epsilon p)^{n-1}}{1- (1-p)^n}$$ is a decreasing function of $p$ for $\epsilon > g(n)$ for some function $g()$ and in particular 
$$\frac{p (1- \epsilon p)^{n-1}}{1- (1-p)^n} \geq (1-\epsilon)^{n-1}$$ for all $\epsilon > g(n)$ with equality at $p=1$. I tried to show that the derivative with respect to p is negative, however without success. Does anyone have any idea how to show this otherwise?
 A: The expression:
$$ \frac{p (1- \epsilon p)^{n-1}}{1- (1-p)^n} $$
is well-defined for $p \in (0,1]$, integer $n \ge 2$ and $0 \lt \epsilon \lt 1$.
We take as the problem to identify for which values of $\epsilon$ the expression is monotone decreasing.  As Comments on the the Question demonstrate, the behavior of the expression for varying $\epsilon$ values depends on $n$, with some combinations of values producing monotonic decreases and others not.
Note that while the expression as written is undefined for $p=0$, both numerator and denominator tend to zero.  Further, an application of l'Hôpital's Rule shows that $\frac{1-(1-p)^n}{p}$ tends to $n$ as $p\to 0^+$, so the expression above tends to $1/n$ "at" $p=0$.
Thus a necessary condition for this expression to be a decreasing function of $p$ on $(0,1]$ is that:
$$ \frac{1}{n} \ge (1-\epsilon)^{n-1} $$
Solving more explicitly for $\epsilon$ produces the following necessary condition:
$$ \epsilon \ge 1 - \left( \frac{1}{n} \right)^{\frac{1}{n-1}} $$
The rest of our post provides a sufficient condition, and we establish this by taking the derivative and invoking Descartes Rule of Signs.
For convenience we rewrite the expression at top in terms of $q = 1 - p$, so that our inquiry is about whether:
$$ F(q) \equiv \frac{(1-q)(1 - \epsilon(1-q))^{n-1}}{1-q^n} $$
is increasing as a function of $q$ on $[0,1)$.
Cancelling $1-q$ from numerator and denominator simplifies $F(q)$ and allows us to evaluate $F(0)$ and $F(1)$ without recourse to l'Hôpital's Rule:
$$ F(q) = \frac{(\epsilon q + 1 -\epsilon)^{n-1}}{\sum_{k=0}^{n-1} q^k} $$
This function is continuous on $[0,1]$ and by inspection $F(0)=(1-\epsilon)^{n-1}$ and $F(1)=1/n$.  Our previous observation about the necessary condition is now reframed as a requirement for increasing, $F(0) \le F(1)$.
The derivative with respect to $q$ is a straightforward application of the quotient rule:
$$ \begin{align*} F'(q)
&= \frac{(n-1)\epsilon(\epsilon q + 1 -\epsilon)^{n-2} \sum_{k=0}^{n-1} q^k 
 - (\epsilon q + 1 -\epsilon)^{n-1} \sum_{k=1}^{n-1} k q^{k-1}}{\left( \sum_{k=0}^{n-1} q^k \right)^2} \\
&= \frac{(\epsilon q + 1 -\epsilon)^{n-2}
 \left[ (n-1)\epsilon\sum_{k=0}^{n-1} q^k 
\; - \; (\epsilon q + 1 -\epsilon)\sum_{k=0}^{n-2} (k+1) q^k \right]}{\left( \sum_{k=0}^{n-1} q^k \right)^2} \\
&= \frac{(\epsilon q + 1 -\epsilon)^{n-2}
 \left[ (n-1)\epsilon - 1 +\epsilon
+ \left( \sum_{k=1}^{n-2} ((n-1)\epsilon - (1 - \epsilon)(k+1) - \epsilon k)  q^k \right)
+ (n-1)\epsilon q^{n-1} - (n-1)\epsilon q^{n-1} \right]}{\left( \sum_{k=0}^{n-1} q^k \right)^2} \\
&= \frac{(\epsilon q + 1 -\epsilon)^{n-2}
 \left[ \sum_{k=0}^{n-2} (n\epsilon-1-k) q^k \right]}{\left( \sum_{k=0}^{n-1} q^k \right)^2}
\end{align*} $$
Because the base factor in the numerator $\epsilon q + (1-\epsilon)$ is positive and the denominator is the square of a positive term, the sign of derivative $F'(q)$ is determined by the sign of the degree $n-2$ polynomial factor in the numerator:
$$ P(q) = \sum_{k=0}^{n-2} (n\epsilon-1-k) q^k $$
Plainly $F(q)$ is not constant on $[0,1]$ and hence is increasing there if and only if $P(q) \ge 0$ on $[0,1]$.  The coefficients of $P(q)$ are monotone decreasing, so by Descartes Rule of Signs $P(q)$ can have at most one positive real root.
Take note of these computations:
$$ \begin{align*} 
P(0) &= n\epsilon - 1 \\
P(1) &= (n-1)(n\epsilon - 1) - \sum_{k=0}^{n-2} k \\
     &= (n-1)\left( n\epsilon - 1 - \frac{n-2}{2} \right) \\
     &= (n-1)\left( n\epsilon - \frac{n}{2} \right) 
\end{align*}$$
Since $P(1) \ge 0 \iff \epsilon \ge 1/2$, recalling $n \ge 2$, it follows that $\epsilon \ge 1/2$ is both a necessary and sufficient condition for $F(q)$ to be increasing on $[0,1]$ and correspondingly for the original expression to be decreasing on $(0,1]$ as a function of $p$.
