Asymptotic for a binomial sum Is it possible to derive an asymptotic expression as $n \to \infty$ for the following sum:
$$S_n(a,x) = \sum_{k=1}^n \binom{n}{k} \frac{k!^2 x^k}{k \cdot (a)_k} $$
$(a)_k$ is the rising factorial (Pochhammer symbol).
Euler-Maclaurin seems way too complicated here, and I'm not sure what else to try.

Not an answer (so I still hope for someone to post a better one and get the bounty), but I need somewhere to put my notes on this.
Using @Maxim 's suggestion in the comments, we change the summation index.
Denote $k=n-l$, then
$$S_n(a,x) = n! x^n \sum_{l=0}^{n-1} \frac{(n-l-1)!}{l! ~ (a)_{n-l} ~ x^l}$$
$$(a)_{n-l}=\frac{\Gamma(a+n-l)}{\Gamma(a)}$$
$$S_n(a,x) = \Gamma(a) n! x^n \sum_{l=0}^{n-1} \frac{\Gamma(n-l)}{l! ~ \Gamma(n-l+a) ~ x^l}$$
$$S_n(a,x) = n! x^n \sum_{l=0}^{n-1} \frac{B(n-l,a)}{l! x^l}$$
$$B(n-l,a)=\int_0^1 t^{n-l-1}(1-t)^{a-1}dt$$
However, Beta function is not very useful here. Instead, we could use Stirling approximation for Gamma function, which up to 2 orders, reads:
$$\Gamma(n+1)\asymp {\sqrt {2\pi n}}\left({\frac {n}{e}}\right)^{n}\left(1+{\frac {1}{12n}}+{\frac {1}{288n^{2}}}+O(1/n^3) \right)$$
Thus we have for the 0 order approximation, assuming $l$ fixed:
$$\frac{\Gamma(n-l)}{\Gamma(n-l+a)} \asymp e^a  \left(\frac{n-l-1}{n-l-1+a}\right)^{n-l-1/2} \frac{1}{(n-l-1+a)^a} $$
So, we have asymptotic series:
$$S_n(a,x) \asymp \Gamma(a) n^{-a} n! x^n \sum_{l=0}^{m} \frac{1}{l!x^l}$$
Where $m$ is some number, $m\ll n$.
However, I checked numerically for some values, and it's a very poor approximation.
We may need to use more orders of Stirling approximation. I'll return to this problem as soon as I'm able.
