PART I
The following series, according to W. Mathematica, does converge to
$$\sum_{k = 1}^{+\infty} \frac{e^{-k}}{k^k \sqrt{k}} = 0.3929049383779132(...)$$
The previous result can be written in terms of elementary numbers plus the Euler constant, as follows:
$$0.3929049383779132 \approx \frac{-92-95 e+86 e^2}{2 \left(130-4 e+33 e^2\right)}$$
Question Is that conjecture true? Probably it's a lack of mine but I cannot manage to make W. Mathematica to spit out more digits of the previous number.
PART II
The previous series was actually a special case for $x = 1$ of the more general series:
$$\sum_{k = 1}^{+\infty} \frac{x^k e^{-k}}{k^k \sqrt{kx}}$$
Question: Is there a close form for this?
I tried with many values of $x$ and the series always gets a numerical result, but Mathematical cannot give me a close form.
I am not assuming a priori it does exist, but many times people found out close forms whereas software could not.
It is a pure approximation but in closed form.
Using Stirlings formule for $n!\approx\sqrt{2\pi n}\big(\frac{n}{e}\big)^n$ we get:
$\sum_{k = 1}^{+\infty} \frac{x^k e^{-k}}{k^k \sqrt{{k}{x}}}$$\approx$$\sqrt\frac{{2\pi}}{x}$$\sum_{k = 1}^{+\infty} \frac{({xe^{-2}})^{k}}{k!}$=$\sqrt\frac{{2\pi}}{x}$$\big(e^{xe^{-2}}-1\big)$
At $x=1$ its value is about $0.363262....$