# Simplification of Gamma

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

• $\lambda\frac{k}{\alpha-1}$
– robjohn
Mar 24, 2018 at 6:53
• @robjohn how??? Mar 24, 2018 at 6:53
• $\Gamma(k+1)=k\,\Gamma(k)$
– robjohn
Mar 24, 2018 at 6:54

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 $$\lambda\frac{\Gamma(\alpha-1)}{\Gamma(\alpha)}\cdot\frac{\Gamma(k+1)}{\Gamma(k)}=\lambda\frac{k}{\alpha-1}$$
Use that $$\Gamma (n) =(n-1)!$$
Hence it simplifies to $$\lambda\cdot\frac {(\alpha-2)!}{(\alpha-1)!}\cdot\frac {k! }{(k-1)!}$$
• @Fabian Why not? $$(\alpha-1)!=(\alpha-1)(\alpha-2)!$$ and $$k! =k(k-1)!$$ This simplifying the expression further to $$\frac {k\lambda}{\alpha-1}$$ Mar 24, 2018 at 7:00