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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?

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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

1 Answer 1

up vote 2 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 $\Gamma(a+1)=a\Gamma(a)$.

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