# Expressing $\prod_{k=1}^n \left( k - \frac{1}{2} \right)$ using the gamma function

I want to express $$\prod_{k=1}^n \left( k - \frac{1}{2} \right)$$ using the gamma function. I think this is equivalent to $\left(k-\frac{1}{2}\right)!$ so I set $a=k-1$ and then used the identity $$\Gamma \left(n+\frac{1}{2}\right) = {(2n)! \over 4^n n!} \sqrt{\pi}$$ to get $$\prod_{k=1}^n \left( k - \frac{1}{2} \right) = \left(a+\frac{1}{2}\right)! = {\Gamma(2k - 1) \over 4^{k-1} \Gamma(k)} \sqrt{\pi}$$ However, Wolfram Alpha disagrees with this much more succinct answer: $$\frac{\Gamma\left(k+\frac{1}{2}\right)}{\sqrt{\pi}}$$

Wolfram Alpha's answer looks more like the integer relation between gamma and factorial... except for the factor of $\sqrt{\pi}$ that I'm at a loss to explain. How do I get from my finite product to this expression?

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${\large\Gamma\left(n + 1/2\right)/\sqrt{\,\pi\,}\,}$. – Felix Marin Mar 20 '14 at 22:03

Define $a_n:=\prod_{k=1}^n\left(k-\frac 12\right)$. From the relationship $\Gamma(x+1)=x\Gamma(x)$ for $x$ positive, we derive $$k-\frac 12=\frac{\Gamma(k+1-1/2)}{\Gamma(k-1/2)},$$ hence the product which defines $a_n$ is telescopic. We obtain $$a_n=\frac{\Gamma(n+1/2)}{\Gamma(1/2)}.$$ Since $\Gamma(1/2)=\sqrt\pi$, we get the same formula as Wolfram Alpha.

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Ugh, I was so overwhelmed by the other answers that I missed yours which is succinct and to the point (and essentially what I wrote). Another deletion for me. (+1) – robjohn Mar 20 '14 at 22:37
I was just reviewing some old deleted answers and noticed that mine was different. I believe that the correct answer is $$\frac{\Gamma(n+1/2)}{\sqrt\pi}$$ From the rest of your answer, your formula for $a_n$ looks like a typo. – robjohn Feb 13 '15 at 14:10
You are right, I fixed the typo, thanks. Unfortunately, it is now the same as your deleted answer, so I am not sure I deserve the reputation points; – Davide Giraudo Feb 13 '15 at 14:40
Your answer was first, and I don't hold typos against anyone. You deserve the points. Don't spend them all in one place ;-) – robjohn Feb 13 '15 at 15:22

The first type of identity needs no gamma function at all to arrive at a correct result. $$\prod_{k=1}^n\left(k-\tfrac12\right)=\frac1{2^n}\prod_{k=1}^n(2k-1) =\frac1{4^nn!}\prod_{k=1}^n(2k)(2k-1)=\frac{(2n)!}{4^nn!}$$

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The factorial has two equivalent definitions. First, the one I expect you have learnt

$k!=k\cdot(k-1)\ldots2\cdot1$ or, "$k$ factorial is the product of all the integers from $1$ up to $k$"

This is a good definition in that it's natural - it seems sensible given the situations in which the factorial is helpful (in combinatorics, for instance). On the other hand, it does not lend itself to sensible generalisations. There is an equivalent definition, though, in the form of a recurrence relation.

$k!=k\cdot (k-1)!$ , and $1!=1$

Now, this is obviously the same as the first definition when $k$ is a positive integer, but if we define a sensible value for, say, $(1/2)!$, then we can generate values of $(3/2)!$, $(5/2)!$, etc.

What you have done is set $(1/2)!=1/2$ (and, I would guess, $x!=x$ for all $x \in (0,1)$) and that's fine, but it turns out that, while this seems like a sensible generalisation, it isn't in fact the most interesting or natural one. Let's call your generalisation $f(x)$, and define it on the positive reals as $$f(x) = xf(x-1) \text{ and }f(x)=x \text{ if } x \in (0,1]$$ We have that $f(x)=x!$ when $x$ is a positive integer, and we can now put in any positive real we like. It disagrees with our usual convention of $0!=1$ when $x$ is near $0$, but that's fine too - we don't necessarily have to agree with that convention.

The Gamma function, though, is a little different. $\Gamma(x)$ is defined usually as $$\Gamma(x)=\int_{0}^{\infty}u^{x-1}e^{-u}du$$ and it can indeed be shown that $\Gamma(x+1)=x\Gamma(x)$ and $\Gamma(1)=1$, and therefore that $\Gamma(x+1)=x!$ when $x$ is a positive integer (which is a slight translation of our second definition for the factorial). However, $\Gamma(x)$ has a completely different set of values in $[0,1]$ than our $f(x)$ (including, strangely, that $\Gamma(1/2)=\sqrt{\pi}$). The reason this is usually the definition is that it comes up pretty much everywhere.

So the reason your two answers look similar is because any function satisfying the recurrence relation above will have a value at $n+1/2$ of that sort of form, but the specific value at $1/2$ is what introduced the $\sqrt{\pi}$.

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