Simplifying a Gamma over a Gamma

Question

I am in the final stages of a proof and need help. I have simplified my starting expression expression down to $\dfrac{v\Gamma(-1+\frac{v}{2})}{2\Gamma(\frac{v}{2})}$

I know the above expression is to equal $\frac{v}{v-2}$

I am having ahard time getting there.

I know $\Gamma(n) = (n-1)!$

So I thought something like $\dfrac{v*(\dfrac{v}{2}-2)(\dfrac{v}{2}-1)(\dfrac{v}{2}-0)}{2(\dfrac{v}{2}-1)(\dfrac{v}{2}-0)}$ Would work but the terms only cancel to produce $\dfrac{v*(\dfrac{v}{2}-2)}{2}$

Can anyone give me a push in the right direction?

Answer 1

Use $\Gamma(z+1) = z \Gamma(z)$.

See https://en.wikipedia.org/wiki/Gamma_function#Main_definition.

Answer 2

$\Gamma\left(\frac{v}{2}\right) = \left(\frac{v}{2}-1\right)\,\Gamma\left(\frac{v}{2}-1\right)$ hence $$ \frac{v\,\Gamma\left(\frac{v}{2}-1\right)}{2\,\Gamma\left(\frac{v}{2}\right)}= \frac{v\,\Gamma\left(\frac{v}{2}-1\right)}{2\,\left(\frac{v}{2}-1\right)\,\Gamma\left(\frac{v}{2}-1\right)}=\frac{v}{2\left(\frac{v}{2}-1\right)}=\frac{v}{v-2}.$$