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?