# $(\frac{2 \times 2}{1 \times 3}) \times (\frac{4 \times 4}{3 \times 5}) \times (\frac{6 \times 6}{5 \times 7}) \times \cdots = \frac{\pi}{2}$

I'm having a bit of trouble converting one infinite product to another... I'm sure I'm just missing something obvious somewhere, but any help would be greatly appreciated.

I've gotten stuck on a certain part of the text: "Leonhard Euler's Integral: A Historical Profile of the Gamma Function" by Philip J. Davis, which can be found online here. From pg 853:

Apparently, Euler, experimenting with infinite products of numbers, chanced to notice that if $$n$$ > is a positive number,

$$[(\frac{2}{1})^n \times \frac{1}{n+1}] \times [(\frac{3}{2})^n \times \frac{2}{n+2}] \times [(\frac{4}{3})^n \times \frac{3}{n+3}] \times \cdots = n!$$

Leaving aside all delicate questions as to the convergence of the infinite product, the reader can verify this equation by cancelling out all the common factors which appear in the top and bottom of the left-hand side. Moreover, the left-hand side is defined (at least formally) for all kinds of of $$n$$ other than negative integers. Euler noticed also that when $$n = \frac{1}{2}$$ is inserted, the left-hand side yields (after a bit of manipulation) the famous infinite product of the Englishman John Wallis (1616-1703):

$$(\frac{2 \times 2}{1 \times 3}) \times (\frac{4 \times 4}{3 \times 5}) \times (\frac{6 \times 6}{5 \times 7}) \times (\frac{8 \times 8}{7 \times 9}) \times \cdots = \frac{\pi}{2}$$

I can verify to myself that the first equation yields $$n!$$ for positive integers of n, but I'm struggling a bit to convince myself that Euler's first infinite product is equal to Wallis's second infinite product for $$n = \frac{1}{2}$$. When I try to rearrange the terms myself, I'm getting something like:

$$\frac{2}{3} \times \frac{2 \times 2}{5} \times \frac{3 \times 2}{7} \times \frac{4 \times 2}{7} \times \cdots$$

Also, empirically, when I take the product of the first 1000 terms or so using a computer, I get a value around 0.886, rather than the stated $$\frac{\pi}{2}$$.

Could someone assist me with the algebraic manipulations needed to convert the first infinite product into the second one? Not necessarily looking for a formal proof, just trying to gain a little intuition.

You just have the wrong value for the factorial:$$\tfrac12!=\Gamma(\tfrac32)=\tfrac12\Gamma(\tfrac12)=\tfrac12\sqrt{\pi}\approx0.8862.$$Similarly, the $$n=\tfrac12$$ case isn't the Wallis product because the first factor is actually $$\sqrt{\tfrac21}\tfrac23$$.