# Prove properties of the gamma function

$$\sum_{n=1}^{∞} \frac{\Gamma(n)}{\Gamma(n+p+1)}=\frac{1}{p^2\Gamma(p)}$$

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Please do not vandalize your question. –  Zev Chonoles Feb 3 '14 at 19:45

Here is an approach. Recalling the function $\frac{1}{1-x}$, then the series in consideration can related to it by noticing it is the fraction integral of order $p+1$

$$\sum_{n=1}^{∞} \frac{\Gamma(n)}{\Gamma(n+p+1)}x^n=\sum_{n=0}^{∞} \frac{\Gamma(n+1)}{\Gamma(n+p+2)}x^{n+p+1}=\frac{1}{\Gamma(p+1)}\int_{0}^{x}(x-t)^{p}\frac{1}{1-t}dt.$$

Now, just evaluate the last integral and then substitute $x=1$. See a related problem.

Note: It is easier to substitute $x=1$ in the integral and then evaluate it, that's is

$$\frac{1}{\Gamma(p+1)}\int_{0}^{1}(1-t)^{p}\frac{1}{1-t}dt =\dots\,.$$

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The Beta function to the rescue:

$$\frac{\Gamma(n)\Gamma(p+1)}{\Gamma(n+p+1)} = B(n,p+1) = \int_0^1 t^{n-1}(1-t)^p\,dt.$$

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+1 That's all. The simplest one. –  Felix Marin Feb 3 '14 at 19:48
How can I induce $\sum_{n=1}^∞ \int_{0}^{1} t^{n-1}(1-t)^p dt=\frac{1}{p}$? –  Delete Feb 3 '14 at 21:11
@Delete By the monotone convergence theorem, $$\sum_{n=1}^\infty \frac{\Gamma(n)\Gamma(p+1)}{\Gamma(n+p+1)} = \sum_{n=1}^\infty \int_0^1 t^{n-1}(1-t)^p\,dt = \int_0^1 \left(\sum_{n=1}^\infty t^{n-1}\right)(1-t)^p\,dt.$$ You know the sum of a geometric series, and that gives you a simple integral whose value is $\frac{1}{p}$. –  Daniel Fischer Feb 3 '14 at 21:15
Thanks a lot. :) –  Delete Feb 3 '14 at 21:16

Have you looked at the function $$F_{p}(x)=\sum_{n=1}^{\infty}\frac{x^{n+p}}{n(n+1)\cdots(n+p)},\;\;\; |x| < 1 ?$$ The value you want is $\lim_{x\uparrow 1}F_{p}(x)$.

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