Hypergeometric function with negative $b$ and $a>c>0$ Recall the definition of the hypergeometric function 
$$_2F_1(a,b,c;z)=\sum_{n=0}^{\infty}\frac{(a)_n(b)_n}{n!(c )_n}x^n$$
where $(a)_n$ is defined to be $a(a+1)\cdots(a+n-1)$.

We suppose that none of the $a,b,c$ or the difference between any two of them is an integer.

Here, I am interested in the case where $a&gt;c&gt;0,b&lt;0$.

What's the behaviour of this function on $[0,1]$?
I used Mathmatica to plot it and it suggests that it always goes negative somewhere on $[0,1]$. As $_2F_1(a,b,c;0)=1$, this suggests that the function has at least a zero on $[0,1]$.

Is there anyway to prove it?
$_2F_1(a,b;c;z)$

$=(1-z)^{-a}{}_2F_1\left(a,c-b;c;\dfrac{z}{z-1}\right)$ (according to http://en.wikipedia.org/wiki/Hypergeometric_function#Transformation_formulas)

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(a)_n(c-b)_nz^n}{(c)_nn!(1-z)^{n+a}}$
$_2F_1(a,b;c;z)$

$=(1-z)^{-a}{}_2F_1\left(a,c-b;c;\dfrac{z}{z-1}\right)$ (according to http://en.wikipedia.org/wiki/Hypergeometric_function#Transformation_formulas)

$=\sum\limits_{n=0}^\infty\dfrac{(-1)^n(a)_n(c-b)_nz^n}{(c)_nn!(1-z)^{n+a}}$