# Evaluate the limit without using Stirling Formula

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

• I doubt this is possible, because $pi$ pops up in the answer. Maybe you can show that Stirling's formula follows from this limit? Although that might also be unlikely, because $e$ does not pop up... Mar 9 '19 at 23:57
• I was thinking of using the Walli's Product which seems a good starting point but I can't get anywhere near yet @SmileyCraft Mar 10 '19 at 0:11
• Interesting idea. I believe Wallis' product can only be used to solve $p=2$, though. Mar 10 '19 at 0:43
• Is there a typo in your first limit? I think the current limit evaluates to $p^{\tfrac32-p}(2\pi)^{\tfrac{p-1}{2}}$. The second limit (in your edit) does equal the right hand side of the first equation. Mar 10 '19 at 19:34
• Unless the book has a typo it should be correct. The limit is exatcly $p\Gamma(1/p) \Gamma(2/p)\cdots \Gamma(p/p)$ using that $\Gamma(x)=\lim_{n\to \infty} \frac{n^x n!}{(x(x+1)\cdots(x+n)}$ @Servaes Mar 10 '19 at 22:57

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:

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

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

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

• Could you give some reference about the product develompent of sine? Mar 11 '19 at 12:29
• @JoseSquare : There is endless information about it, please look for "sine product expansion" or "sine product formula" or "Eulers infinite product for the sine". The formula is $\Gamma(x)\Gamma(1-x)=\pi/\sin(\pi x)$ , see Wikipedia. Old literature e.g. here Mar 11 '19 at 12:41
• @JoseSquare : Or e.g. Wolfram Mathworld, here . Mar 11 '19 at 12:52
• Okey I knew the Eulers infinite product, I thought you meant a product formula like $\prod_{k=1}^{p-1} \sin (\frac{\pi k}{p}) = p 2^{1-p}$. Your answer is really satisfactory to me because in the book is shown the relation between sine and Gamma without using Stirling's formula. I'll wait a few days with the bounty still open just in case another answer is given. However I don't think there is a simpler way. Mar 11 '19 at 13:04
• @JoseSquare : The product $\displaystyle\prod\limits_{k=1}^{p-1}\sin\frac{\pi k}{p}=p2^{1-p}$ comes from $\displaystyle\frac{\sin px}{\sin x}=2^{p-1}\prod\limits_{k=1}^{p-1}\sin\left(x+\frac{\pi k}{p}\right)$ (where $x\to 0$), and more general it's a result of $\displaystyle\frac{a^n-b^n}{a-b}=\prod\limits_{k=1}^{n-1}(a-be^{i2\pi k/n})$ . --- Sorry, I don't know yet where we can find a proof for free. (Not for free: "Formulas and Theorems in Mathematics" written by George S.Carr.) Mar 11 '19 at 13:54

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 $$a$$.

• Interesting idea. I let the question still open wating if anybody has another way to aproach the problem. Thanks for your effort Mar 10 '19 at 23:04