Property of the Gamma function

Question

I want to know is there any property due to which we could write $$\frac{\Gamma(M+\frac{2}{\alpha})}{\Gamma(M)}=\frac{2}{\alpha}\sum_{k=1}^{M} \frac{\Gamma(k-1+\frac{2}{\alpha})}{(k-1)!}$$ Thanks in advance.

Answer 1

Use $\Gamma(x+1) = x \cdot \Gamma(x)$ and induction. If $M=1$ then $$\frac{\Gamma(1+\frac{2}{\alpha})}{\Gamma(1)}=\Gamma\left(1+\frac{2}{\alpha}\right) =\frac{2}{\alpha}\Gamma\left(\frac{2}{\alpha}\right)=\frac{2}{\alpha}\sum_{k=1}^{1} \frac{\Gamma(k-1+\frac{2}{\alpha})}{(k-1)!}.$$

If $M>1$ then $$\frac{2}{\alpha}\sum_{k=1}^{M} \frac{\Gamma(k-1+\frac{2}{\alpha})}{(k-1)!}= \frac{2}{\alpha}\sum_{k=1}^{M-1} \frac{\Gamma(k-1+\frac{2}{\alpha})}{(k-1)!}+ \frac{2}{\alpha}\frac{\Gamma(M-1+\frac{2}{\alpha})}{(M-1)!}\\= \frac{\Gamma(M-1+\frac{2}{\alpha})}{\Gamma(M-1)}+\frac{2}{\alpha}\frac{\Gamma(M-1+\frac{2}{\alpha})}{(M-1)!}=\left(M-1+\frac{2}{\alpha}\right)\frac{\Gamma(M-1+\frac{2}{\alpha})}{(M-1)!}=\frac{\Gamma(M+\frac{2}{\alpha})}{\Gamma(M)},$$ where we used the induction step and the fact that $\Gamma(n) = (n-1)!$ for any positive integer $n$.