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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} -1k^{*}? \end{equation}

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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 $|r|a\frac{1}{a}$. 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

enter image description here

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    It is a very good idea to put everything in a closed form but I am not able to see easily that $f(a)$ is increasing after all. What do you mean by dependence on the inverse of $a$?2012-04-27
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    @SeyhmusGüngören: The idea is quite simple here. If you have a number $01$. Besides, if $K>k^*$, $\frac{1}{a^K}>\frac{1}{a^{k^*}}$. In this case, the hypergeometric function does help and your function, with the given intervals, is seen to increase. Also note that, in your case, $a|r|< a$ and so the numerator gives a larger number than the denominator.2012-04-27
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    yes but there is a difference term inbetween and the other gamma and hypergeometric terms are also confusing as what happens finally. I am also surprised that it is simply correct for you while I am still trying to understand what is going on)))2012-04-27
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    @SeyhmusGüngören: I have improved the answer. I hope this will help you.2012-04-27
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    Could you fill in some of the details of your answer? I don't think "hypergeometric functions are also helpful in this direction" could be more vague.2012-04-27
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    I accept that nominator is always greater than the denominator and nominator and the denominator have the same signs. These two information do not suffice to deduce that $f(a)$ is increasing with $a$. I can also have the same information from the original description. If the derivative could be shown to be greater than zero than I would say okay. In its current version I cannot understand any improvement.2012-04-27
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    @AntonioVargas: I will provide more thorough information, thanks for your comment. For Seyhmus: I think you are right and I am in need for a better clarification. Of course, having a formula in a closed form is far better than the ratio of sums!2012-04-28
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    My friend is interested in how you found the very first formula with $\Gamma$s thanks alot.2012-04-30
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    @SeyhmusGüngören: Indeed, your sums can be given in a closed form. This can be obtained using Wolfram Alpha or Mathematica. The fact that appears Gamma products is just due to the fact that one does not assume $k^*$ and $K$ integers otherwise these are just $\binom{K}{k^*}$. It is interesting to note that the reason why a hypergeometric function appears is that your sums do not run to infinity otherwise one can us the Newton binomial formula and end the story.2012-04-30
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    thank you very much for the explanation!2012-04-30