Simplification of Gamma

Question

Simplify

$$\lambda\frac{\Gamma(\alpha-1)}{\Gamma(\alpha)}\cdot\frac{\Gamma(k+1)}{\Gamma(k)}$$

My attempt,

Since $\Gamma(\alpha)=(\alpha-1)\Gamma(\alpha-1)$

So the expression becomes $$\frac{\lambda}{\alpha-1}\cdot\frac{\Gamma(k+1)}{\Gamma(k)}$$.

How to proceed? Thanks a lot

Answer 1

Do the same thing you did with $\Gamma(\alpha)=(\alpha-1)\,\Gamma(\alpha-1)$. Substitute $\alpha=k+1$ instead and you get $\Gamma(k+1)=k\,\Gamma(k)$. Substitute this into the expression given and you get $$\lambda\frac{\Gamma(\alpha-1)}{\Gamma(\alpha)}\cdot\frac{\Gamma(k+1)}{\Gamma(k)}=\lambda\frac{k}{\alpha-1}$$

Answer 2

Hint

Use that $$\Gamma (n) =(n-1)!$$

Hence it simplifies to $$\lambda\cdot\frac {(\alpha-2)!}{(\alpha-1)!}\cdot\frac {k! }{(k-1)!}$$