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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

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  • $\begingroup$ $\lambda\frac{k}{\alpha-1}$ $\endgroup$
    – robjohn ♦
    Mar 24, 2018 at 6:53
  • $\begingroup$ @robjohn how??? $\endgroup$
    – Mathxx
    Mar 24, 2018 at 6:53
  • 2
    $\begingroup$ $\Gamma(k+1)=k\,\Gamma(k)$ $\endgroup$
    – robjohn ♦
    Mar 24, 2018 at 6:54

2 Answers 2

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

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Hint

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

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

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  • $\begingroup$ Does not look simpler to me... $\endgroup$
    – Fabian
    Mar 24, 2018 at 6:58
  • $\begingroup$ @Fabian Why not? $$(\alpha-1)!=(\alpha-1)(\alpha-2)!$$ and $$k! =k(k-1)!$$ This simplifying the expression further to $$\frac {k\lambda}{\alpha-1}$$ $\endgroup$
    – Rohan Shinde
    Mar 24, 2018 at 7:00

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