TITLE: infinite sum of harmonic number powers $\sum_{n=0}^{\infty}x^{H_n}$
QUESTION [5 upvotes]: I am looking for a simplified form of the infinite sum of harmonic exponentials
$$
f(x) = \sum_{n=0}^{\infty}x^{H_n}
= 1 + x + x^{3/2} + x^{11/6} + x^{25/12} + x^{137/60} + x^{49/20} + \ldots\\
=1+x\left(
1+x^{1/2}\left(
1+x^{1/3}\left(
1+x^{1/4}\left(
1+x^{1/5}\left(
1+x^{1/6}\left(
1+\ldots
\right)\right)\right)\right)\right)\right)
$$
for $x\in[0,1/e)$.
Can it be reduced to a standard special function?
Update: a stable way of calculating $f(x)$ numerically
Inspired by Claude Leibovici's answer, we write
$$
f(x) = 1 + \sum_{n=1}^{\infty}x^{\gamma+\log(n)} - \underbrace{\sum_{n=1}^{\infty}\left[x^{\gamma+\log(n)}-x^{H_n}\right]}_{g(x)}\\
= 1+x^{\gamma}\zeta[-\log(x)]-g(x)
$$
with $g(x)$ converging rapidly on the entire interval $x\in[0,1/e]$ (and beyond) without singularities:

Specifically,
$$
g(1/e) = \sum_{n=1}^{\infty}\left[\frac{e^{-\gamma}}{n}-e^{-H_n}\right]
\approx 0.35370447459102146589\ldots\\
g'(1/e) \approx 0.38530115152397563963\ldots\\
g''(1/e) \approx -0.82751608018475931083\ldots
$$
give the approximate series expansion of $f(x)$ around $x=1/e$,
$$
f(x) \approx
\frac{e^{-\gamma-1}}{1/e-x}
+ 0.36556578362553594918\ldots
+ (1/e-x)\cdot0.11500038261356357678\ldots
+ (1/e-x)^2\cdot0.11786065053337882776\ldots
+O[(1/e-x)^3]
$$
Unfortunately, neither WolframAlpha nor AskConstants recognize any of these numbers.

REPLY [4 votes]: As @Raffaele commented, if you use the approximation
$$H_n\sim \log( n)+\gamma$$
Using what you wrote (notice that $H_7=2.59$ while $\log(7)+\gamma=2.52$)
$$f(x)\sim1 + x + x^{3/2} + x^{11/6} + x^{25/12} + x^{137/60} + x^{49/20} +
x^\gamma\sum_{n=7}^{\infty}n^{\log(x)}$$
$$\sum_{n=7}^{\infty}n^{\log(x)}=\zeta (-\log (x))-\sum_{k=1}^6 k^{\log(x)}$$ which converges for $x\lt \frac 1e$.
Let us compare for a few values of $x$
$$\left(
\begin{array}{ccc}
 x & \text{approximation} & \text{exact} \\
 0.00 & 1.00000 & 1.00000 \\
 0.02 & 1.02424 & 1.02421 \\
 0.04 & 1.05408 & 1.05391 \\
 0.06 & 1.08957 & 1.08911 \\
 0.08 & 1.13148 & 1.13055 \\
 0.10 & 1.18100 & 1.17937 \\
 0.12 & 1.23976 & 1.23718 \\
 0.14 & 1.30998 & 1.30621 \\
 0.16 & 1.39482 & 1.38955 \\
 0.18 & 1.49874 & 1.49168 \\
 0.20 & 1.62838 & 1.61920 \\
 0.22 & 1.79396 & 1.78233 \\
 0.24 & 2.01210 & 1.99766 \\
 0.26 & 2.31175 & 2.29413 \\
 0.28 & 2.74822 & 2.72704 \\
 0.30 & 3.44199 & 3.41684 \\
 0.32 & 4.71483 & 4.68530 \\
 0.34 & 7.81187 & 7.77753 \\
 0.36 & 26.6198 & 26.5802
\end{array}
\right)$$ which seems to be quite decent.
So, you can write
$$\sum_{n=0}^{\infty}x^{H_n}=\sum_{n=0}^{m}x^{H_n}+x^\gamma\Big[\zeta (-\log (x))-\sum_{k=1}^{m+1} k^{\log(x)}\Big]\tag 1$$
As @Roman commented, if you take a large value of $m$, the asymptotics is
$$f(x)\sim  \sum_{n=0}^{m}x^{H_n}-\frac{m\, x^{\gamma+\log (m)}}{1+\log(x)}$$
Edit
Equation $(1)$ can be simplified ; in terms of the  Hurwitz zeta function, it is
$$f(x)=\sum_{n=0}^{\infty}x^{H_n}=\sum_{n=0}^{m}x^{H_n}+x^\gamma \,\zeta (-\log (x),m+1)$$ and expanded as series
$$f(x)=\sum_{n=0}^{m}x^{H_n}-x^{\gamma+\log(m)} g(t)\qquad \text{where} \qquad t=\log(x)$$
$$g(t)=\frac{m}{t+1}+\frac{1}{2}+\frac{t}{12 m}-\frac{(t-2) (t-1) t}{720 m^3}+\frac{(t-4)
   (t-3) (t-2) (t-1) t}{30240 m^5}-\frac{(t-6) (t-5) (t-4) (t-3) (t-2) (t-1)
   t}{1209600 m^7}+O\left(\frac{1}{m^9}\right)$$