Closed form for $\sum_{n=1}^{\infty}\frac{x^n}{n!\sqrt{n}}$, or an asymptotic for it

Question

In my study I came about the function defined by

$$f(x)=\sum_{n=1}^{\infty}\frac{x^n}{n!\sqrt{n}}$$

and I have not been able to find any sort of closed form for it, nor have I been able to find any sort of assymptotics as $x\to+\infty$. Plotting the function on Desmos I can conjecture that

$$f(x)\sim c\frac{e^x}{\sqrt{x}}$$

for some constant $c$ (possibly equal to 1), but I cannot prove it. In general, is there a way to express the function

$$g(x)=\sum_{n=1}^{\infty}\frac{x^n}{n!n^{\alpha}}$$

where $\alpha\in\mathbb{R}>0$? This function seems to be some sort of exponential analogue of the poly-logarithm $\mathrm{Li}_s(n)$ defined by

$$\mathrm{Li}_s(n)=\sum_{n=1}^{\infty}\frac{x^n}{n^{\alpha}}$$

which does not have any closed form (unless $\alpha$ is a negative integer). If one could express $f(x)$ in terms of $\mathrm{Li}_{1/2}(n)$ then that would be perfect.

Answer 1

The Laplace transform of your function $g$ is $$\mathscr{L} g(s) = \int_0^\infty \sum_{n=1}^\infty \frac{x^n}{n! n^\alpha} \exp(-sx)\; dx = \sum_{n=1}^\infty \frac{s^{-n-1}}{n^\alpha} = \frac{\text{Li}_\alpha(1/s)}{s}$$ for $s > 1$.