Bailey's Theorem

Let $\Gamma(z)$ be the Gamma Function, then

$\left[{\Gamma(m+{\textstyle{1\over 2}})\over\Gamma(m)}\right]^2\underbrace{\lef... ... m+1} +\left({1\cdot 3\over 2\cdot 4}\right)^2 {1\over m+2}+\ldots}\right]}_{n}$

$=\left[{\Gamma(n+{\textstyle{1\over 2}})\over\Gamma(n)}\right]^2\underbrace{\le... ... n+1} +\left({1\cdot 3\over 2\cdot 4}\right)^2{1\over n+2}+\ldots}\right]}_{m}.$

$\left[{\Gamma(m+{\textstyle{1\over 2}})\over\Gamma(m)}\right]^2\underbrace{\lef... ... m+1} +\left({1\cdot 3\over 2\cdot 4}\right)^2 {1\over m+2}+\ldots}\right]}_{n}$
$=\left[{\Gamma(n+{\textstyle{1\over 2}})\over\Gamma(n)}\right]^2\underbrace{\le... ... n+1} +\left({1\cdot 3\over 2\cdot 4}\right)^2{1\over n+2}+\ldots}\right]}_{m}.$