can someone please help me to get the value of the following one? $\sum_{n=1}^{n=\infty}\frac{p^{n}}{\{(a+2n\alpha)(a+(2n-2)\alpha).........(a+2\alpha)\}\{(b+(2n-1)\alpha)(b+(2n-3)\alpha).......(b+\alpha)\}}$
If we can write $(a+2n\alpha)(a+(2n-2)\alpha)......(a+2\alpha)$ and $(b+(2n-1)\alpha)(b+(2n-3)\alpha).......(b+\alpha)$ as a pochhammer symbol, may be we can write it as hypergeometric function.The $1st$ term can be expressed as pochhammer easily but how can I make the $2nd$ one. kindly help.
$$\prod_{k=1}^{n}(a+2k\alpha) = (2\alpha)^n\cdot\left(1+\frac{a}{2\alpha}\right)_n\tag{1}$$ $$\prod_{k=1}^{n}(b+(2k-1)\alpha) = (2\alpha)^n\cdot \left(\frac{b +\alpha}{2\alpha}\right)_n \tag{2}$$ hence your series equals
$$ \sum_{n\geq 1}\frac{\left(\frac{p}{4\alpha^2}\right)^n}{ \left(1+\frac{a}{2\alpha}\right)_n \left(\frac{1}{2}+\frac{b }{2\alpha}\right)_n}\\=\frac{p}{4\alpha^2\Gamma\left(\frac{3}{2}+\frac{b}{2\alpha}\right)\Gamma\left(2+\frac{a}{2\alpha}\right)}\;\phantom{}_{3}F_2\left(1,1,1;2+\frac{a}{2\alpha},\frac{3}{2}+\frac{b}{2\alpha};\frac{p}{4\alpha^2}\right) \tag{3}$$