Evaluate the limit without using Stirling Formula

Question

I want to prove that

$$\lim_{n \to \infty} p \frac{n^{(p+1)/2} (n!)^p p^{np+1}}{(np+p)!} = p^{1/2} (2\pi)^{(p-1)/2}$$

So I am working the book The Gamma Function by Emile Artin. In the book this limit is involved for proving the Gauss multiplication formula of the Gamma Function. The book uses the Stirling Formula for $n!$ to show the limit is that. However I was wondering if there is another way of prooving this limit that doesn't use Stirling Formula. Does anyone sees any way of approaching this??

Edit

The limit making some manipulations is the same as

$$\lim_{n \to \infty} p \frac{(n!)^p p^{np}}{(np)!n^{(p-1)/2}}$$

maybe this helps someone attempmting this task

Answer 1

You can use the product development of sine instead of the Stirling formula.

$$\displaystyle\lim_{n \to \infty} p\frac{(n!)^p p^{np}}{(np)!n^{(p-1)/2}} = p\prod\limits_{k=1}^{p-1}\Gamma\left(\frac{k}{p}\right)= p\sqrt{\prod\limits_{k=1}^{p-1}\Gamma\left(\frac{k}{p}\right)\Gamma\left(1-\frac{k}{p}\right)}=$$

$$\hspace{3.8cm}\displaystyle = p\sqrt{\prod\limits_{k=1}^{p-1}\frac{\pi}{\sin(\frac{\pi k}{p})}} = p\sqrt{\frac{\pi^{p-1}}{2^{1-p}p}}=\sqrt{p(2\pi)^{p-1}}$$

Note:

$$\prod\limits_{k=1}^{p-1}\sin\left(x+\frac{\pi k}{p}\right) = \frac{1}{(i2)^{p-1}} \prod\limits_{k=1}^{p-1}\left( e^{i(x+\frac{\pi k}{p})}- e^{-i(x+\frac{\pi k}{p})}\right)$$

$$= \frac{1}{(i2)^{p-1}} \prod\limits_{k=1}^{p-1} e^{i\frac{\pi k}{p}} \prod\limits_{k=1}^{p-1}\left( e^{ix}- e^{-ix-i2\frac{\pi k}{p}}\right)$$

$$= \frac{1}{(i2)^{p-1}} e^{i\frac{\pi(p-1)}{2}}\frac{e^{ipx}-e^{-ipx}}{e^{ix}-e^{-ix}} = 2^{1-p}\frac{\sin(px)}{\sin x} ~~(\,\to p2^{1-p}\, \text{ for}\, x\to 0\,).$$

Answer 2

From the de Moivre - Laplace theorem, or at least one particular form of it, we have that $$ {an \choose n} \left(\frac{1}{a}\right)^n \left(\frac{a-1}{a}\right)^{(a-1)n} \sim \frac{1}{\sqrt{2 \pi an \frac{1}{a} \left(\frac{a-1}{a}\right)}} \tag{1} $$ as $n \to \infty$, for every positive integer $a$. (The symbol "$\sim$" denotes that the limit of the quotient of left and right sides is 1.) The most direct way to prove this is by using Stirling's formula, but you can prove it without Stirling's formula by more general probability theory. See for example Theorem 7 in Tao's notes on variants of the central limit theorem.

Your equation (as expressed in your edit) is $$ \lim_{n \to \infty} a \frac{(n!)^a a^{an}}{(an)! n^{(a-1)/2}} = a^{1/2} (2\pi)^{(a-1)/2}, $$ which can also be written as $$ \frac{(an)!}{(n!)^a} \sim \sqrt{a} \frac{a^{an}}{(2 \pi n)^{(a-1)/2}}. $$ This is equivalent to equation (1), by using $$ \frac{\frac{(an)!}{(n!)^a}}{\frac{((a-1)n)!}{(n!)^{(a-1)}}} = {an \choose n} $$ and induction on $a$.