Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. Join them; it only takes a minute:

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

We can use the property $\Gamma{(a + 1)} = a \Gamma{(a + 1)}$ to simply the expression below to $\frac{a}{a + b}$ if $k = 1$.

$$ \frac{\Gamma{(a+b)}}{\Gamma{(a)}} \frac{\Gamma(a+1)}{\Gamma{(a + 1 + k)}} $$

How does it simplify? Here's what I've tried.

$$ \frac{\Gamma{(a+b)}}{\Gamma{(a)}} \frac{\Gamma(a+1)}{\Gamma{(a + 1 + k)}} \\ \frac{\Gamma{(a+b)}}{\Gamma{(a)}} \frac{a\Gamma(a)}{\Gamma{(a + 2)}} \\ \frac{\Gamma{(a+b)}}{\Gamma{(a)}} \frac{a\Gamma(a)}{a^2\Gamma{(a)}} \\ \frac{a^b\Gamma{(a)}}{\Gamma{(a)}} \frac{a\Gamma(a)}{a^2\Gamma{(a)}} \\ \frac{a^{b+1}}{a^2} \\ a^{b - 1} $$

That doesn't look like $\frac{a}{a + b}$. Where did I err?

share|cite|improve this question
The fisrt identity is wrong. It should read $a \Gamma(a) = \Gamma(a+1)$. Plus, you are using it incorrectly. $$\Gamma(a+2) = (a+1) \Gamma(a+1) = (a+1)a\Gamma(a)$$ – Pragabhava Nov 15 '12 at 6:23
up vote 3 down vote accepted

My guess is that the formula simplifies to $\frac{a}{a+b}$ when $k=b$ (and not when $k=1$). To see this, use the identities $\Gamma(a+1+b)=(a+b)\Gamma(a+b)$ and $\Gamma(a+1)=a\Gamma(a)$.

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.