Is there a closed form for this sequence?

Question

I'm trying to find a closed form for the following sequence:

$a$

$a(a-1)$

$a(a-1)(a-2)$

$a(a-1)(a-2)(a-3)$

The problem is, $a=\frac{1}{2}$. If it were some whole number, then I'd use

$\frac{a!}{(a-n)!}$

I thought maybe I could re-frame the problem to use the factorial, but I can't figure it out. How do I do this?

Well this is ugly, but works: $$\large\prod_{i=0}^{n} (a-i)$$ Or you can define the factorial like this: $$\large n!=\prod_{i=1}^{n} i$$ and it will work. — Teoc, Jan 11 '15 at 20:21
The Pochhammer symbol should help (you'll get a quotient of $\Gamma$ functions). — Raymond Manzoni, Jan 11 '15 at 20:27

score 0 · Answer 1 · edited Apr 13 '17 at 12:20

For the special case $a=\frac{1}{2}$ we can get a closed form. I didn't give an answer to the general solution however.

Write it down as a product. This gives $$\prod_{k=0}^n (a-k)$$

We'll put $a=\frac{1}{2}$ and see what this gives. By substitution we have $$\prod_{k=0}^n (\frac{1}{2}-k)=\prod_{k=0}^n\frac{1-2k}{2}$$

Now write out the $n$th partial product as follows

$$\prod_{k=0}^n\frac{1-2k}{2}=(\frac{1}{2})(-\frac{1}{2})(-\frac{3}{2})...(-\frac{2n-1}{2})$$

and we can see $$\prod_{k=0}^n\frac{1-2k}{2}=(-1)^n\frac{\prod_{i=1}^n 2i-1}{2^{n+1}}=(-1)^n\frac{(2n!)}{2^n(n!)}\frac{1}{2^{n+1}}=(-1)^n\frac{(2n!)}{2^{2n+1}(n!)}$$

where the product evaluated in the numerator I got from here.

Is there a closed form for this sequence?

1 Answers1