As the title says, I wish to prove the limit (as $n\to \infty$) $$\frac{\left(\Gamma(1 + 1/p)\right)^n}{\Gamma(1 + n/p)}\to 1\qquad \text{ for } p =\frac{\ln n}{\ln\frac n {n-2}}$$ Any hints?
The expression is the volume of the p-ball $B_p(1/2)=\{x:\|x\|_p≤1/2\}$ . The parameter $p$ is chosen so that $B_p(1/2)$ is tangent to the orthoschemes of $[−1/2,1/2]^n$. All p-ball volumes converge to $0$ for fixed $p$. The orthoscheme has volume $1/n!$ by the determinant simplex formula. I expect the volume of the tangent ball to converge to $1$ and numerically it does...
The idea is that a p-ball volume converges to $0$, as well as the orthoschemes of the hypercube, but if $p$ grows appropriately as the dimensions grow, then clearly that is not the case. (For instance, $[-1/2,1/2]^n$ is just $\{x : \|x\|_\infty \leq 1/2\}$). To find such a $p$, one way is to guess the p-ball will be tangent to the orthoschemes.
For the p-ball volume formula see the note [1]. [1] Xianfu Wang, Volumes of Generalized Unit Balls, DOI:10.2307/30044198
I solved it in the following manner: First, I prove that $n/p\to 0$. This comes from: $$n/p = \frac{n\log(\frac n{n-2})}{\log(n)} \to \frac 2 \infty = 0$$ Then the denominator is dealt with, as $\Gamma$ is continuous. As for the numerator, the fact that $n/p\to 0$ along with the fact that $\Gamma$ is analytic around $1$ gives me $\left(\Gamma(1 + 1/p)\right)^n \to 1$ by using the log expansion,
$$n\ln\Gamma(1 + 1/p) = -\gamma n/p + O(n/p^2) \to 0$$ In fact this is also true for p-balls with $p = n^{1 + \epsilon}$.