A finite summation

Let $m,n$ be positive integers. Calculate

$\mathcal{S} = \sum_{k=1}^{2n} \prod_{i=0}^{m} \left ( \left \lfloor \frac{k+1}{2} \right \rfloor + \alpha + i \right )^{-1}$

where $\alpha$ is a non negative number and $\left \lfloor x \right \rfloor$ represents the greatest integer less than or equal to $x$ .

Solution

Using the identity $\Gamma(x+1) = x \Gamma(x)$ we note that

$\prod_{i=0}^{m} \frac{1}{b+i} = \frac{\Gamma(b)}{\Gamma(b+m+1)} = \frac{1}{m} \left ( \frac{\Gamma(b)}{\Gamma(b+m)} - \frac{\Gamma(b+1)}{\Gamma(b+m+1)} \right )$

Hence,

$\begin{align*} \sum_{k=1}^{2n} \prod_{i=0}^{m} \left ( \left \lfloor \frac{k+1}{2} \right \rfloor + \alpha + i \right )^{-1} &= \sum_{k=1,3,\dots, 2n-1} \prod_{i=0}^{m} \left ( \frac{k+1}{2} + \alpha +i \right )^{-1} + \\ & \quad \quad \quad \quad \quad + \sum_{k=2, 4, \dots, 2n} \prod_{i=0}^{m} \left ( \frac{k}{2}+ \alpha + i \right )^{-1} \\ &= 2 \sum_{k=1}^{n} \prod_{i=0}^{m} \left ( k + \alpha + i \right )^{-1} \\ &= \frac{2}{m} \sum_{k=1}^{n} \left ( \frac{\Gamma(\alpha + k)}{\Gamma\left ( \alpha + k +m \right )} - \frac{\Gamma\left ( \alpha + k +1 \right )}{\Gamma \left ( \alpha + k +m +1 \right )} \right ) \\ &= \frac{2}{m} \left ( \frac{\Gamma(\alpha + 1)}{\Gamma\left ( \alpha + 1 +m \right )} - \frac{\Gamma\left ( \alpha + n +1 \right )}{\Gamma \left ( \alpha + n+m +1 \right )} \right ) \end{align*}$