Does division of polynomials give an increasing function?

Question

How can I show that \begin{equation} f(a)=\frac{\sum_{i=1}^{k^*-1} \left(\begin{array}{c} K \\ i \\ \end{array} \right) \left(-1-\frac{1}{ar}\right)^i+1}{\sum_{i=1}^{k^*-1} \left(\begin{array}{c} K \\ i \\ \end{array} \right) \left(-1+\frac{1}{a}\right)^i+1} \end{equation} is an increasing funtion of $a$ for \begin{equation} -1<r<0,\hspace{3mm} 0.5<a<1,\hspace{3mm} \mbox{and}\hspace{3mm} K>k^{*}? \end{equation}

Answer 1

This function can be cast in a closed form as $$ f(a)=\frac{\left(-\frac{1}{ar}\right)^K-\left(-1-\frac{1}{ar}\right)^{k^*}\frac{\Gamma(K+1)\ _2F_1(1,k^*-K,1+k^*,1+\frac{1}{ar})}{\Gamma(k^*+1)\Gamma(K-k^*+1)}}{\left(\frac{1}{a}\right)^K-\left(-1+\frac{1}{a}\right)^{k^*}\frac{\Gamma(K+1)\ _2F_1(1,k^*-K,1+k^*,-1+\frac{1}{a})}{\Gamma(k^*+1)\Gamma(K-k^*+1)}}. $$ The function $_2F_1$ is the hypergeometric function. Now, we note that $a<1$ and $|r|<1$ and so, for $K>k^*$ and the dependence on the inverse of $a$ make this an increasing function in the given intervals.

We recognize a simpler rewriting of this formula as $$ f(a)=\frac{\left(-\frac{1}{ar}\right)^K-\left(-1-\frac{1}{ar}\right)^{k^*}\binom{K}{k^*}\ _2F_1(1,k^*-K,1+k^*,1+\frac{1}{ar})}{\left(\frac{1}{a}\right)^K-\left(-1+\frac{1}{a}\right)^{k^*}\binom{K}{k^*}\ _2F_1(1,k^*-K,1+k^*,-1+\frac{1}{a})}. $$

Now, one has $r<0$ and $|r|a<a$ always. This implies that numerator and denominator have always the same signs. But there is an important difference that is $\frac{1}{a|r|}>\frac{1}{a}$. Hypergeometric functions are also helpful in this direction. This can be easily seen with some numerical check (I did it with Mathematica) and one see that, in the given intervals, it is always $\ _2F_1(1,k^*-K,1+k^*,1+\frac{1}{ar})<\ _2F_1(1,k^*-K,1+k^*,-1+\frac{1}{a})$. I think, but I have not done it, that checking the properties of this function one can avoid a numerical check.

Looking at this formula at increasing $a$, we note at the numerator $a$ appears always multiplied by a reducing factor $r$ that maintains it always smaller than the quantity appearing at the denominator and so, increasing it with the given interval and properly multiplying it for increasing but negative $r$, this function can only increase. For the sake of completeness, I give here a graph of this for $k^*=100$ and $K=150$. For $r=-0.3$ you will get