I wanted to derive
$$\left(\frac{1}{2}\right)! = \frac{\sqrt\pi}{2}$$
from the infinite product definition of the gamma function
$$\Gamma(z+1)=\prod_{n=1}^{\infty}\frac{(1+\frac{1}{n})^z}{1+\frac{z}{n}}$$
On plugging in $\;z=\dfrac{1}{2}\;$ I get
$$\Gamma\left(\frac{3}{2}\right) = \left(\frac{1}{2}\right)! = \frac{\sqrt{1+\frac{1}{1}}\cdot\sqrt{1+\frac{1}{2}}\cdot\sqrt{1+\frac{1}{3}}\dots}{(1+\frac{1}{2})\cdot(1+\frac{1}{4})\cdot(1+\frac{1}{6})\dots} = \frac{\sqrt{1+\frac{1}{1}}\cdot\sqrt{1+\frac{1}{3}}\cdot\sqrt{1+\frac{1}{5}}\dots}{\sqrt{1+\frac{1}{2}}\cdot\sqrt{1+\frac{1}{4}}\cdot\sqrt{1+\frac{1}{6}}\dots} = \sqrt{\frac{2\cdot \frac{4}{3}\cdot \frac{6}{5}\dots}{\frac{3}{2}\cdot \frac{5}{4}\cdot \frac{7}{6} \dots}} = \sqrt{\frac{2^2 \cdot 4^2 \cdot 6^2 \dots}{3^2 \cdot 5^2 \cdot 7^2 \dots}}=\sqrt{\frac{\pi}{2}}$$
according to the Wallis product. For some reason I can't find the mistake, although I suspect it to be really dumb.
Gamma Function – Infinite Product Definition of (1/2)!
factorialgamma functionpi
Best Answer
The mistake is not obvious, because everything you are doing is algebraically correct. There is however a problem with the way you rearrange the product.
Let's show how on a simple case: the Wallis product itself:
$$\frac{\pi}{2} = \Big(\frac{2}{1} \cdot \frac{2}{3}\Big) \cdot \Big(\frac{4}{3} \cdot \frac{4}{5}\Big) \cdot \Big(\frac{6}{5} \cdot \frac{6}{7}\Big) \cdot \Big(\frac{8}{7} \cdot \frac{8}{9}\Big) \cdot \; \cdots$$
Let's rearrange factors a little bit. Here I only remove the factor $1$, and slide every other factor of the denominator. It should be harmless, right?
$$\frac{\pi}{2} \ne \Big(\frac{2}{3} \cdot \frac{2}{3}\Big) \cdot \Big(\frac{4}{5} \cdot \frac{4}{5}\Big) \cdot \Big(\frac{6}{7} \cdot \frac{6}{7}\Big) \cdot \Big(\frac{8}{9} \cdot \frac{8}{9}\Big) \cdot \; \cdots$$
The equality can't hold now, as all factors of the product are now $<1$. The product is actually diverging to $0$.
You did the same "sliding" when you removed $\sqrt{1+\dfrac1{2k}}$ from the numerator.
Besides, the example shows you that your last equality (when you indeed use the Wallis product) is also dubious: as it is written, I may read it as a divergent product as well: it depends on where you put the parens.
It's dangerous to write an infinite product informally as a product of factors ending with "$\dots$", even more so if you write a ratio of two infinite products, with an unspecified order of operations. Order does matter: $\prod_{k=1}^{\infty} u_k$ is defined as the limit of $p_n=\prod_{k=1}^{n} u_k$. If you add things from other factors in $p_n$, it may look like overall the factors are the same (in the unspecified, dots notation), but the limit may not be the same.