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I need to prove that, for $s$ being a nonnegative integer:

$$(-2s-1)!! = \frac{(-1)^s}{(2s-1)!!} = \frac{(-1)^s2^ss!}{(2s)!}$$

where this $!!$ stands for 'double factorial'

Studying the expression $(-2s-1)!!$ I came with:

$$(-2s-1)!! = (-2s-1)(-2s-3)(-2s-5)\cdots(-2s-2n)$$

First of all, shouldn't this product be infinite? Also, I think this wouldn't help, since we must come with a result that is less than $1$. No pure algebraic manipulation would work. I think the gamma function should be used.

See this identity:

$$\Gamma\left(n+\frac{1}{2}\right) = \frac{(2n-1)!!}{2^n}\sqrt{\pi}\implies (2n-1)!! = 2^n\frac{\Gamma\left(n+\frac{1}{2}\right)}{\sqrt{\pi}}$$

Now take $n=-s$ and we have:

$$(-2s-1)!! = 2^{-s}\frac{\Gamma\left(-s+\frac{1}{2}\right)}{\sqrt{\pi}}$$

I still don't know how to express $\Gamma\left(-s+\frac{1}{2}\right)$, and that $\sqrt{\pi}$ must vanish. Any ideas?

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  • $\begingroup$ Isn't the double factorial for a negative odd number just defined to be $(-1)^s/(2s-1)!!$? $\endgroup$ – Vim May 10 '17 at 23:38
  • $\begingroup$ Are you interest in a proof for the integers, reals, or complex numbers? $\endgroup$ – Michael McGovern May 10 '17 at 23:40

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