Express $a_n=\frac{1.5.9.13...(4n+1)}{2^n}$ in terms of gamma function

Question

$a_n=\frac{1.5.9.13...(4n+1)}{2^n}$

I've tried albeit unsuccessfully to generate gamma functions that will be equal to the above.

Tried using the gamma function of fractions but didn't achieve anything with it:

$$\Gamma(\frac{2n-1}{2})=[\frac{(2n-3)(2n-5)...1}{2^{n-1}}]\pi$$

Who can help?

score 1 · Accepted Answer · answered Oct 07 '15 at 20:05

1

Try $\Gamma({13\over4})$, or generally, $\Gamma(n+{1\over4})$. The problem is, you will end up with $\Gamma({1\over4})$ and won't be able to get rid of it. That's how things work.

answered Oct 07 '15 at 20:05

Ivan Neretin

12,835

You also have to multiple that answer by $2^n$. And actually, you've gotten the off-by-one part of gamma wrong - it needs to be $\Gamma(17/4)$ to work. – Thomas Andrews Oct 07 '15 at 20:06
Yes, my fault. It's $\Gamma(n+{5\over4})$ times something, then. – Ivan Neretin Oct 07 '15 at 20:09
Thanks guys. You've made my day. – Obinoscopy Oct 07 '15 at 20:27

Express $a_n=\frac{1.5.9.13...(4n+1)}{2^n}$ in terms of gamma function

1 Answers1