(1/2)! from the infinite product definition of gamma

Question

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.

Answer 1

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.

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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.

Answer 2

Let us consider $$a_k = \prod_{n=1}^{k} \sqrt{\frac{2n}{(2n-1)(2n+1)}}$$

You erroneously noted: $$\left(\frac{1}{2}\right)! = \lim_{k\to \infty} a_k$$

And then you made a mistake in evaluating it: $$\lim_{k\to \infty} a_k = \lim_{k \to \infty} \frac{\sqrt{\pi}}{\sqrt{2}}\cdot \prod_{n=1}^k \frac{1}{\sqrt{2n}}=0$$ so this product on its own evaluates to $0$ rather than anything close to Wallis product. The error occurred due to the ambiguities posed by the $\dots$ notation. In order to avoid such issues you should ideally use the $\Pi$ notation. Always take care on how you deal with factors in infinite products.

Instead to find $\Gamma\left(\frac{3}{2}\right)$ using Wallis product and the multiplicative formula, you can go about as follows:

Noting that $\Gamma(z+1)=z\Gamma(z)$

$$\Gamma(1+z)\cdot\Gamma(-z) = \frac{\Gamma(1+z)\cdot \Gamma(1-z)}{-z} = -\frac{1}{z}\prod_{n=1}^{\infty} \frac{1}{1-\frac{z^2}{n^2}} = -\frac{1}{z}\prod_{n=1}^{\infty} \frac{n^2}{n^2-z^2}$$ The last two equalities hold due to the formula you used, $\Gamma(1+z) = \prod_{n=1}^{\infty} \frac{(1+\frac{1}{n})^z}{1+\frac{z}{n}}$.

In particular when $z=-\frac{1}{2}$, it means $$\Gamma^2\left(\frac{1}{2}\right) = 2\prod_{n=1}^{\infty} \frac{4n^2}{(2n-1)(2n+1)} = 2\cdot\frac{\pi}{2}$$ where the last product is exactly Walli's product. Therefore $$\Gamma\left(\frac{1}{2}\right) = \sqrt{\pi}$$

Now, via the first functional equation,

$$\Gamma\left(\frac{3}{2}\right) = \frac{\Gamma\left(\frac{1}{2}\right)}{2}$$