# gamma and beta function relation

I am reading this, and don't understand how this is reached $$\frac{\Gamma(\alpha)}{\Gamma(\alpha+n)}=\frac{(\alpha+n)\beta(\alpha+1,n)}{\alpha\Gamma(n)}$$

The following relation mentioned on the Wikipedia,$$\beta(\alpha,n)=\frac{\Gamma(\alpha)\Gamma(n)}{\Gamma(\alpha+n)} \Rightarrow \frac{\Gamma(\alpha)}{\Gamma(\alpha+n)}=\frac{\beta(\alpha,n)}{\Gamma(n)}$$

so given the above two, this must be true, $$\frac{(\alpha+n)\beta(\alpha+1,n)}{\alpha\Gamma(n)}=\frac{\beta(\alpha,n)}{\Gamma(n)}$$ which means this also must be true, $$\frac{(\alpha+n)\beta(\alpha+1,n)}{\alpha}=\beta(\alpha,n)$$. I can't see why this is true, given the definition of the $\beta$ function.

• Use the fact that $\Gamma(n)=(n-1)\Gamma(n-1)$ on the expanded $\beta(\alpha+1,n)$ Jul 26 '16 at 15:26
• This could be on topic on the Mathematics SE site, but is also on topic here, IMO. We can defer to the OP's preferences. Jul 26 '16 at 16:13

The last statement you have \begin{equation} \frac{(\alpha+n)\beta(\alpha+1,n)}{\alpha} = \beta(\alpha,n) \end{equation} is true if you expand $\beta(\alpha+1,n)$ as $\frac{\Gamma(\alpha+1)\Gamma(n)}{\Gamma(\alpha+1+n)}$ then follow @ArtificialBreeze's suggestion. Here it is
Cancelling out terms will result in $\frac{\Gamma(\alpha)\Gamma(n)}{\Gamma(\alpha+n)}$, which is what you have above.