# Proof that $\prod\limits_{r=1}^{\infty}\Gamma(\tfrac{1}{3}r)=\frac{640}{3^6}\left(\frac{\pi}{\sqrt{3}}\right)^3$

This problem is from Whittaker & Watson's chapter on the Gamma function, apparently taken from a Peterhouse examination.

Show that $$\prod\limits_{r=1}^{\infty}\Gamma\left(\tfrac{1}{3}r\right)=\frac{640}{3^6}\left(\frac{\pi}{\sqrt{3}}\right)^3$$

Intuitively, because I can show easily that $$\Gamma\left(\tfrac{1}{3}\right)\Gamma\left(\tfrac{2}{3}\right)=\frac{2\pi}{\sqrt{3}}$$

and since all the Gamma functions with other non-integer arguments in the product can be related back to this product just by scaling by a factor $>1,$ (e.g. I can take pairs of terms 4 & 5, 7 & 8, etc. and use the relation $\Gamma(z+1)=z\Gamma(z)$ to get $\Gamma\left(\tfrac{4}{3}\right)\Gamma\left(\tfrac{5}{3}\right)=\tfrac{4\pi}{9\sqrt{3}},$ etc.) it seems that this product can't converge in the usual sense, because the general term won't tend to $1$ as $r\to\infty.$ Wolfram Alpha also believes that this product should not converge.

So it seems that to get this result, some other definition of convergence may be at work here. I am reluctant to simply assume that this is 'wrong' and dismiss the question, because the quality of the problems in this book is generally of a very high standard, and I have finished almost all the other problems, so I want to complete it fully.

Can somebody suggest to me some way that I can justify (even if not 'rigorously' - I don't care so much about rigour here) the above result? I have considered some kind of clever rearrangement, but I'm not sure how to proceed with this, or if this really is the right way to go about it. A full answer is not necessary, but a helping hand would be much appreciated.

Thanks.

• just invert your infinity and take $8$ :) you get finite product with necessary result – serg_1 Jul 28 '17 at 13:12
• Great, just a typo then. Thanks – Hobbyist Jul 28 '17 at 13:33