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Let $z\in \{z\in\mathbb{C}:|z|<1\}, \alpha>-1,\Gamma(s)$ is the gamma function.

How to prove $\sum_{n\geq 0}{\frac{\Gamma(n+2+\alpha)}{n!\Gamma(2+\alpha)}z^n}=\frac{1}{(1-z)^{2+\alpha}}$ ?

If $\alpha=0,$ then it is easy to get the above result. How can i prove the above general result?

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David Lee
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    it is not true. do you know the Taylor expansion of $(1-z)^{\beta}$ around $z=0$ ? (when $\beta \not \in \mathbb{Z}$) – reuns Apr 04 '16 at 02:47
  • take a look at the binomial_series – reuns Apr 04 '16 at 02:50
  • $$\Gamma(n+2+\alpha) = \Gamma( n-1 + 2 + \alpha)(n+1+\alpha)=\cdots = \Gamma(n-n+2+\alpha)(2+\alpha)\cdots (n+1+\alpha)$$

    So now we have

    $$\sum_{n=0}^\infty \frac{(n+1+\alpha)^{(n)}}{n!} z^n$$ (underline in the notation of wikipedia) What you want for the pruported RHS would be $\binom{-2}{n}$ instead.

     – AHusain Apr 04 '16 at 03:10
  • @[email protected] you. There is a type-error of problem. I re-edited my problem and I think I can fix it by your methods. – David Lee Apr 04 '16 at 03:18

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