How to determine the value of $\displaystyle f(x) = \sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n$?

Question

How to determine the value of $\displaystyle f(x) = \sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n$? No context, this is just a curiosity o'mine.

Yes, I am aware there is no reason to believe a random power series will have a closed form in terms of well established functions, but also I have no way to know if that is the case here, so that is why I'm asking. Do you know this power series or any method I could use to determine its value?

In my research I've found out about the polylogarithm, which is defined as $$\mathrm{Li}_s(x) = \sum_{n=1}^\infty\frac{x^n}{n^s} = \frac1{\Gamma(s)}\int_0^\infty\frac{t^{s-1}}{e^t/x-1}dt$$

This called my attention because $$\begin{aligned} f(x) &= \sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n\\ &= x\sum_{n=1}^\infty\frac1{\sqrt n}\frac{x^{n-1}}{(n-1)!}\\ &= x\sum_{n=1}^\infty\frac1{\sqrt n}\mathcal L^{-1}\left\{\frac1{x^n}\right\}\\ &= x\mathcal L^{-1}\left\{\sum_{n=1}^\infty\frac1{\sqrt n}\frac1{x^n}\right\}\\ &= x\mathcal L^{-1}\left\{\mathrm{Li}_{1/2}\left(\frac 1x\right)\right\} \end{aligned}$$

Yeah... This is not the closed form I was expecting. Can we do better?

---

A comment suggested this post might have the answer to my question. However, despite it asking for the same thing, it asks as well for similar and more general expressions, which motivates less specific responses. In fact, the accepted (and only) answer in the post discusses only the asymptotic behavior of the series I'm interested in.

Answer 1

In my opinion, getting the integral representation is too little for a post, but in order to put some meat on the bones, we can find a full asymptotics using this formula. $$S(x)=\sum_{n=1}^\infty\frac{\sqrt n}{n!}x^n=x\sum_{n=0}^\infty \frac{x^n}{n!\sqrt {n+1}}=x\sum_{n=0}^\infty \frac{x^n}{n!}\frac1{\sqrt\pi}\int_0^\infty t^{-1/2}e^{-(n+1)t}dt$$ Performing summation first, $$S(x)=\frac x{\sqrt\pi}\int_0^\infty e^{xe^{-t}-t}\frac{dt}{\sqrt t}$$ The asymptotics at $x\to0$ is evident. Decomposing $e^{xe^{-t}}$ near $x=0$ $$S(x)\sim\frac x{\sqrt\pi}\int_0^\infty e^{-t}\big(1+xe^{-t}+\frac{x^2}{2!}e^{-2t}+...\big)\frac{dt}{\sqrt t}$$ what, of course, after integration simply coincides with the initial sum.

At $x\to\infty$ $$S(x)\overset{s=e^{-t}}{=}\frac x{\sqrt\pi}\int_0^1\frac{e^{xt}}{\sqrt{\ln\frac1t}}dt\overset{t=1-s}{=}\frac{xe^x}{\sqrt\pi}\int_0^1\frac{e^{-xs}}{\sqrt{\ln\frac1{1-s}}}ds\overset{xs=t}{=}\frac{e^x}{\sqrt\pi}\int_0^x\frac{e^{-t}}{\sqrt{\ln\frac1{1-\frac tx}}}dt$$ The integrand is declining when $t$ is growing, and becomes exponentially small, for example, at $t\sim\sqrt x$. It means that we are allowed to decompose the denominator near $t=0$ and integrate term by term. Expanding integration to $\infty$ and dropping exponentially small terms $$S(x)\sim\frac{e^x}{\sqrt\pi}\int_0^x\frac{e^{-t}}{\sqrt{\frac tx+\frac{t^2}{2x^2}+...}}dt\sim\frac{\sqrt x\,e^x}{\sqrt\pi}\int_0^\infty e^{-t}\Big(1-\frac t{4x}-\frac7{96}\frac1{x^2}+...\Big)\frac{dt}{\sqrt t}$$ $$S(x)=\sqrt x\,e^x\left(1-\frac1{8x}-\frac7{128x^2}+O\Big(\frac1{x^3}\Big)\right)$$

Answer 2

This is not a complete solution but it gives only the result: a complete formula for a sum generalizing that of the OP.

The development was initiated by the results of @Svyatoslav and completed to this point in a discssion with him.

$$s(a,x) := \sum_{n=0}^{\infty} \frac{n^a}{n!}x^n \\ = \frac{x}{\Gamma(1-\{a\})} \frac{\partial ^{\lfloor a\rfloor}}{\partial w^{\lfloor a\rfloor}}\left(\int_0^1 \frac{\exp \left(-u x\; e^w +e^w x+w\right)}{\log ^{\{a\}}\left(\frac{1}{1-u}\right)} \, du\right)|_ {w\to 0}$$

Here $\lfloor a\rfloor$ is the integer part of $a$ and $\{a\} = a-\lfloor a\rfloor$ is the fractional part of $a$.