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I have trouble with the following problem.

Let $f=f(p)$, $p>1$, $0<f<1$, and $\lim_{p\to\infty}f(p)=1$. Find $f(p)$, such that $$\lim_{p\to\infty}p\left(1-\sqrt{p f(p)}\frac{\Gamma(p+1)}{\Gamma(p+\frac 32)}\right)=\frac{3}{20}$$

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How come this question cannot be edited? – smanoos Dec 18 '11 at 19:16
David: do you know a simple equivalent of $\Gamma(p+1)/\Gamma(p+\frac32)$ when $p\to+\infty$? – Did Dec 18 '11 at 20:10
Thank you very much! – David Dec 18 '11 at 22:08
up vote 3 down vote accepted

For positive half-integers, the gamma function is given by $$ \Gamma(\frac{1}{2} + n) = \frac{(2n)!}{4^nn!}\sqrt{\pi}, $$ and so the ratio in question is $$ \frac{\Gamma(p+1)}{\Gamma(p+\frac{3}{2})} = \frac{4^{p+1}p!(p+1)!}{(2p+2)!\sqrt{\pi}} $$ for integral $p$. As $p\rightarrow\infty$, this is asymptotic to $p^{-1/2}$, so the limit in the problem becomes $$ \lim_{p\rightarrow\infty}p\left(1-\sqrt{f(p)}\right). $$ To make this take on a particular finite value $\alpha$, the expression $1-\sqrt{f(p)}$ must be asymptotic to $\alpha/p$, hence $\sqrt{f(p)}$ must be asymptotic to $1-\alpha/p$, hence $f(p)$ must be asymptotic to $1-2\alpha/p$. Setting $\alpha=3/20$, we have $$ f(p) = 1-\frac{3}{10p} $$ as one function that satisfies the problem.

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Sorry, but I got that the ratio $\Gamma (p+1)/\Gamma(p+\frac 32)$ is asymptotic to $\sqrt p$, not to $\frac{1}{\sqrt p}$ – David Dec 19 '11 at 16:28

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