I'm trying to show that the following function is monotonically increasing with respective to p, for $p \in [0, 1]$: $$f(p) = \sum_{k=2}^{n} (-1)^k {{n-1}\choose{k - 1}} \frac{(k-1)!}{\prod_{l = 1}^k (l + \gamma)} \frac{(\gamma+1)p^{k-1}}{ {}_{2}F_{1}(1, 1-k; \gamma+2; p)} = \sum_{k=2}^{n} (-1)^k \frac{(n-1)!}{(n-k)!} \frac{(\gamma+1)!}{(\gamma+k)!} \frac{p^{k-1}}{ {}_{2}F_{1}(1, 1-k; \gamma+2; p)}$$ where ${}_{2}F_{1}(a,b;c;z)$ is the hypergeometric function. What I currently know is that:
- The function $f(p)$ is bounded between 0 and 1, i.e., $0 \leq f(p) \leq 1$.
- The term $\frac{p^{k-1}}{ {}_{2}F_{1}(1, 1-k; \gamma+2; p)}$ is monotonically increasing with $p$, and equals to 0 when $p = 0$, equals to $\frac{\gamma+k}{\gamma+1}$ when $p = 1$.
The reason I believe the result is true is based on the following numerical simulation: Simulation with $n \in \{2, ..., 24\} $ and $\gamma = 1$. I have tried other $\gamma$s, the simulation results are similar. And I have tried many ways to prove the claim, include taking the derivatives w.r.t $p$ (which is really messy, and cannot see if it is positive) and recognizing that $f(p)$ is of bounded variation. However, none of them seem to work. I have stuck on this problem for about a month now, at this moment, any help would be appreciated! Thank you so much!
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I am aware that there could be more background materials with this question, so let me re-format the question to something that potentially more clear. Just wondering if anyone know how to prove the following inequality:
$$\sum_{k=2}^{n} (-1)^k \frac{(n-1)!}{(n-k)!} \frac{(\gamma+1)!}{(\gamma+k)!} \frac{p^{k-1}}{ {}_{2}F_{1}(1, 1-k; \gamma+2; p)} \leq $$ $$\sum_{k=2}^{n} (-1)^k \frac{(n-1)!}{(n-k)!} \frac{(\gamma+1)!}{(\gamma+k)!} \frac{p^{k-1}(\gamma+k)}{(\gamma+1)}$$ where I have replaced the hypergeometric function with its minimum value (when $p = 1$). Without the $(-1)^k$ term in $f(p)$, the function is monotonically increasing with $p$. However, with the $(-1)^k$ term in $f(p)$, I am not sure how to prove the $\leq$ sign. Again, any help will be appreciated! Thanks!