# How to show that $\prod\limits_{r=1}^{n}\Gamma{ \left({\frac {r}{n+1}}\right)}={\sqrt {\frac {(2\pi )^{n}}{n+1}}}$?

I saw the following equation on Wikipedia, but I am not sure how to approach it.

$$\prod_{r=1}^{n}\Gamma{ \left({\frac {r}{n+1}}\right)}={\sqrt {\frac {(2\pi )^{n}}{n+1}}}$$

Here are some other values listed on Wikipedia

$$\prod _{r=1}^{2}\Gamma \left({\tfrac {r}{3}}\right)={\frac {2\pi }{\sqrt {3}}}$$ $$\prod _{r=1}^{3}\Gamma \left({\tfrac {r}{4}}\right)={\sqrt {2\pi ^{3}}}$$ $$\prod _{r=1}^{4}\Gamma \left({\tfrac {r}{5}}\right)={\frac {4\pi ^{2}}{\sqrt {5}}}$$ Should I use induction for this problem? Any hints on where I should start on?

• It might be of help to take a look at Gauss Multiplication Formula for which a proof can be found within the given link. – mrtaurho Dec 20 '18 at 22:12
• The formula $\Gamma(1-z)\Gamma(z) = \frac{\pi}{\sin(\pi\cdot z)}$ should be useful. Just consider when $n$ is odd and when it's even, and use the formula. – Jakobian Dec 20 '18 at 22:13
• If you square both sides you see that $$\prod_{r=1}^{n}\Gamma^2\bigg(\frac{r}{1+n}\bigg)=\frac{(2\pi)^n}{1+n}$$ Each $\Gamma^2$ term can be put into the form $$\Gamma\bigg(\frac{2r}{n+1}\bigg)B\bigg(\frac{r}{1+n},\frac{r}{1+n}\bigg)$$ After that I have no idea – clathratus Dec 20 '18 at 22:15

Consider Gauss Multiplication Formula in its general form

$$\forall z \notin \left\{{-\frac m n: m \in \mathbb N}\right\}: \prod_{k = 0}^{n - 1} \Gamma \left(z + \frac k n\right) = (2 \pi)^{(n - 1) / 2 }n^{1/2 - n z} \Gamma(n z)\tag1$$

Now by plugging in $$z=1$$ and $$n=n+1$$ we further get

\begin{align} \prod_{k = 0}^{(n+1)-1} \Gamma \left(1 + \frac k{(n+1)}\right) &= (2 \pi)^{((n+1) - 1) / 2} (n+1)^{1/2 - (n+1)\cdot 1} \Gamma((n+1)\cdot1)\\ \prod_{k = 1}^{n} \frac k{n+1}\Gamma \left(\frac k{n+1}\right) &= \sqrt{\frac{(2 \pi)^{n}}{n+1}} \frac{\Gamma(n+1)}{(n+1)^{n}}\\ \prod_{k = 1}^{n} \frac k{n+1}\prod_{k = 1}^{n}\Gamma \left(\frac k{n+1}\right) &= \sqrt{\frac{(2 \pi)^{n}}{n+1}} \frac{\Gamma(n+1)}{(n+1)^{n}}\\ \frac{n!}{(n+1)^n}\prod_{k = 1}^{n}\Gamma \left(\frac k{n+1}\right) &= \sqrt{\frac{(2 \pi)^{n}}{n+1}} \frac{\Gamma(n+1)}{(n+1)^{n}} \end{align}

$$\therefore~\prod_{k = 1}^{n}\Gamma \left(\frac k{n+1}\right) = \sqrt{\frac{(2 \pi)^{n}}{n+1}}$$

We started by utilizing the fundamental functional property of the Gamma Function $$($$note that the index $$k$$ has been uplifted to starting by $$k=1$$ due the fact that within the original product the first term $$\Gamma(1)$$ equals $$1$$ afterall and is so of minor matter$$)$$, followed by splitting up the product, rewriting the first subproduct in finite terms, and finally making use of the fact that $$n!=\Gamma(n+1)$$ for all $$n\in\mathbb N$$ .

• +1. It was really fine. – Felix Marin Dec 22 '18 at 16:16
• @FelixMarin Thank you kindly. – mrtaurho Dec 22 '18 at 18:02

Given the reflection formula for the $$\Gamma$$ function, $$\Gamma(s)\Gamma(1-s)=\frac{\pi}{\sin(\pi s)}$$, the problem immediately boils down to showing that $$\prod_{k=1}^{n}\sin\left(\frac{\pi k}{n+1}\right) = \frac{2(n+1)}{2^{n+1}},$$ which is well-known. One may use roots of unity or Chebyshev polynomials of the second kind, together with Vieta's formulas.

• I guess the RHS is inversed since a product of sines is bounded by $1$. – yultan Jan 9 at 9:30
• @yultan: of course, now fixed. – Jack D'Aurizio Jan 9 at 22:32