Series with Gamma Function

Question

I am trying to evaluate the following series: \begin{equation*} S_{l, d}=\sum_{k=0}^\infty \frac{1}{(2k)!}\frac{\Gamma\left(k+\frac{1}{2}\right)}{\Gamma(k+l+\frac{d}{2})} \end{equation*} where $\Gamma(.)$ is the Gamma function, $d \in \mathbb{R}$, $l \in \mathbb{N}$. I tried using $\Gamma(k+\frac{1}{2})=\frac{(2k)!}{2^{2k}k!} \sqrt{\pi}$ but without success because of $\frac{1}{\Gamma(k+l+\frac{d}{2})}$. Do you have any idea on on how to get a better expression of $S_{l, d}$ ?

Answer 1

Depends on what you need out of it, of course, but one option is to recognise it as a series for the $I$-Bessel function, from $$ I_\alpha(x) = \sum_{k = 0}^\infty \frac{(x / 2)^{2 k + \alpha}}{k! \Gamma(k + \alpha + 1)}. $$

Using the expression for $\Gamma(k + 1/2)$ above, we have $$ S_{l, d} = \sum_{k = 0}^\infty \frac{\sqrt{\pi}}{\Gamma(k + l + d / 2) 4^k k!}, $$ and so taking $x = 1$ we have $$ S_{l, d} = 2^{d / 2 + l - 1} \sqrt{\pi} I_{d/2 + l - 1}(1). $$