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Consider the classical hypergeometric function $F(5/4,3/4; 2, z)$ for $z\in (0,1]$. Is this bounded by some real number (independent of $z$)?

I'm aware of Euler's formula:

$F(5/4,3/4; 2, z) = \frac{1}{\Gamma(5/4)\Gamma(3/4)}\int_0^1 t^{-1/4} (1-t)^1/4 (1-tz)^{-5/4} dt.$

The best I can do using this formula is $F(5/4, 3/4; 2, z) \leq \frac{4}{ 3\Gamma(5/4) \Gamma(3/4)}\frac{1}{1-z}. $ This is not good though, because it blows up as $z$ tends to $1$.

Any suggestions?

Maybe it is easy to show that $F$ is strictly increasing on $(0,1]$ and continuous on $\mathbf{R}$. Then, we simply need to estimate $F(5/4,3/4;2,1)$.

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    haha. you're right. :)2012-03-10

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It is defined for $z \in [-1,1)$, but it diverges to $+\infty$ as $z \to 1^-$.

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All terms in the power series are nonnegative, so the function is increasing on $[0,1)$. Actually it is increasing on $[-1,1)$. I get numerically $F(5/4,3/4;2;1/4) \approx 1.1409$ as the bound for $[0,1/4]$.

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    That's all I need. Thanks.2012-03-10