Given $a>b>0$, is there any upper bound of the following ratio of [hypergeometric function](https://en.wikipedia.org/wiki/Hypergeometric_distribution)?
$$\frac{\_2F\_1(a,1-b;a+1;x)}{\_2F\_1(a,1-b;a+1;y)}$$
for $1>x>y>0$ ideally in the form like some powers of $x/y$？



Ratio of hypergeometric function
If $b$ is not an integer, this ratio is bounded by a constant.


Indeed, Remark 1.2.4 pp 34-35 of "From Gauss to Painlevé" by Iwasaki-Kimura-Shimomura-Yoshida can be applied to your case, in which $\gamma-\alpha-\beta=b$. It yields that the function $f={}\_2F\_1(\alpha,\beta;\gamma,\cdot)$ you consider is continuous on [0,1]. the exact value at $1$ is given on page 73 of the same book at as $\frac{\Gamma(a+1)\Gamma(b)}{\Gamma(1)\Gamma(a+b)}$, in particular it is not $0$.
As f(0)=1, using the local uniqueness of solutions in $(0,1)$, one may conclude the existence of $K>0$ such that $f(x)>K$, for every $x\in[0,1]$.


Then the function $F:[0,1]\times [0,1]\rightarrow \mathbb R, (x,y)\mapsto f(x)/f(y)$ is continuous on a compact, whence the conclusion.


Edit: Actually, the derivative $f'$ is $\left(\frac{a+1}{a(1-b)}\right){}\_2F\_1(\alpha+1,\beta+1;\gamma+1,\cdot)$ and must have the sign of $\frac{a+1}{a(1-b)}$.
So an explicit upper bound can be given:


* if $1-b>0$ then $F$ is bounded by $f(1)/f(0)=\frac{\Gamma(a+1)\Gamma(b)}{\Gamma(1)\Gamma(a+b)}$,
* if $1-b<0$ then $F$ is bounded by $f(0)/f(1)=\left(\frac{\Gamma(a+1)\Gamma(b)}{\Gamma(1)\Gamma(a+b)}\right)^{-1}$.


