Find function if known some limit

Question

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 $$\lim_{p\to\infty}p\left(1-\sqrt{p f(p)}\frac{\Gamma(p+1)}{\Gamma(p+\frac 32)}\right)=\frac{3}{20}$$

Answer 1

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.