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A hypergeometric series

Let \alpha, \beta \in \mathbb{R} such that 0<\alpha<\beta. Evaluate the series:

    \[\mathcal{S} = \sum_{n=0}^{\infty} \frac{\alpha \left ( \alpha+1 \right ) \cdots \left ( \alpha+n \right )}{\left (\beta+1 \right )\left ( \beta+2 \right )\cdots\left ( \beta+n \right )}\]

Solution

Lemma 1: For the \mathrm{B}, \Gamma functions , it holds that:

    \[\mathrm{B}(x, y) = \int_{0}^{1} t^{x-1} \left ( 1-t \right )^{y-1}\, \mathrm{d}t = \frac{\Gamma(x) \Gamma(y)}{\Gamma(x+y)}\]

Lemma 2: It holds that:

    \[\frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \beta+n+2 \right )}= \frac{1}{\Gamma\left ( \beta-\alpha+1 \right )} \cdot \mathrm{B} \left ( n+\alpha+1, \beta-\alpha+1 \right )\]

Proof: Simple calculations using Lemma (1) reveal the identity.

Lemma 3: Using Lemma 2 it holds that

    \[\mathrm{B} \left ( n+\alpha+1, \beta-\alpha+1 \right ) = \frac{1}{\Gamma\left ( \beta-\alpha+1 \right )} \cdot \int_{0}^{1} t^{n+\alpha} \left ( 1-t \right )^{\beta-\alpha} \, \mathrm{d}t\]

and as a consequence

    \[\frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \beta+n+2 \right )} = \frac{1}{\Gamma\left ( \beta-\alpha+1 \right )} \cdot   \int_{0}^{1} t^{n+\alpha} \left ( 1-t \right )^{\beta-\alpha} \, \mathrm{d}t\]

 

Then, successively we have that:

\begin{aligned} \sum_{n=0}^{\infty} \frac{\alpha \left ( \alpha+1 \right ) \cdots \left ( \alpha+n \right )}{\left ( \beta+1 \right )\left ( \beta+2 \right )\cdots \left ( \beta+n \right )} &= \sum_{n=0}^{\infty} \frac{\alpha \left ( \alpha+1 \right )_n}{\left ( \beta+1 \right )_n} \\ &=\sum_{n=0}^{\infty} \frac{\frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \alpha \right )}}{\frac{\Gamma\left ( \beta+n+2 \right )}{\Gamma\left ( \beta \right )}} \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right )} \cdot \sum_{n=0}^{\infty} \frac{\Gamma\left ( \alpha+n+1 \right )}{\Gamma\left ( \beta+n+2 \right )}\\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \sum_{n=0}^{\infty} \mathrm{B} \left ( n+\alpha+1, \beta-\alpha+1 \right )\\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \sum_{n=0}^{\infty} \int_{0}^{1} t^{n+\alpha} \left ( 1-t \right )^{\beta-\alpha} \, \mathrm{d}t \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)}\int_{0}^{1} t^\alpha \left ( 1-t \right )^{\beta-\alpha} \sum_{n=0}^{\infty} t^n \, \mathrm{d}t \\ &=\frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \int_{0}^{1} t^\alpha \left ( 1-t \right )^{\beta-\alpha} \cdot \frac{\mathrm{d}t}{1-t} \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \int_{0}^{1} t^\alpha \left ( 1-t \right )^{\beta-\alpha-1} \, \mathrm{d}t \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \mathrm{B} \left ( \alpha+1, \beta-\alpha \right ) \\ &=\frac{\Gamma\left ( \beta \right )}{\Gamma\left (\alpha \right ) \cdot \Gamma \left(\beta-\alpha+1 \right)} \cdot \frac{\Gamma\left ( \alpha+1 \right ) \Gamma\left ( \beta-\alpha \right )}{\Gamma\left ( \beta+1 \right )} \\ &= \frac{\Gamma\left ( \beta \right )}{\Gamma\left ( \alpha \right ) \cdot \left( \beta -\alpha \right) \cdot \Gamma \left(\beta-\alpha \right)} \cdot \frac{\alpha \Gamma \left (\alpha \right ) \Gamma\left ( \beta - \alpha \right )}{\beta \Gamma \left ( \beta \right )} \\ &= \frac{\alpha}{\beta} \cdot \frac{1}{\beta-\alpha} \end{aligned}

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