Is anything known about the series $\sum_{n=0}^{\infty} x^{\sqrt{n}} $? It's well known that there are a shocking number of identities for the usual Jacobi theta function $$ \theta_3(x) = \sum_{n=-\infty}^{\infty} x^{n^2}. $$
So I wanted to turn my attention to slowly decreasing exponents. If I make my $n$ decay too slow such as $$f(x) = \sum_{n=1}^{\infty} x^{\log n}, $$
we basically have the Riemann zeta function and end up in well explored territory via the relation
$$f(x) = \sum_{n=1}^{\infty} x^{\log n}  = \zeta( - \log x). $$
So then I started to consider some slowly growing exponents that don't grow TOO slowly and the obvious first candidate is
$$ G(x) = \sum_{n=0}^{\infty} x^{\sqrt{n}}. $$
Has anyone looked into this object/similar algebraic exponents? Is it known to obey any interesting identities? It seems like a natural object to consider even without an application/motivation.
 A: In my former research group, we worked, for basic practical applications, the more general problem of
$$G_a(x)=\sum_{n=0}^\infty x^{n^{\frac{1}{a}}}$$ and, for sure, we made the approximation
$$G_a(x)\sim \int_0^\infty x^{n^{\frac{1}{a}}}\,\mathrm d n=\frac{\Gamma(1+a)}{(-\log x)^a}$$ which, for our needs, was more than sufficient.
I would be interested to know if, currently, we could have a better approximation.
A: I claim that the function $G$ satisfies a few functional equations, but in order to formulate our functional equation, we need to extend $G$ to a much larger domain. One often has to extend a function to a larger domain in order for the functional equation to make sense; this is certainly the case for the functional equations involving the Gamma function since one has to extend the factorial function from $\mathbb{N}$ to $\mathbb{C}$ for such a functional equation to be sensible.
Let $H:\{z\in\mathbb{C}\mid\text{Re}(z)<0\}\rightarrow\mathbb{C}$ be the function defined by $H(z)=\sum_{k=1}^{\infty}e^{\sqrt{k}z}$. Then $H(z)=G(e^z)-1$.
Now, let $F$ be the collection of all square free positive integers. Then
$$H(z)=\sum_{k\in F}\sum_{n=1}^{\infty}e^{\sqrt{k}nz}=\sum_{k\in F}\frac{e^{\sqrt{k}z}}{1-e^{\sqrt{k}z}}=\sum_{k\in F}\frac{-1}{2}\cdot e^{\sqrt{k}z/2}\cdot\text{sech}(\sqrt{k}z/2).$$
Generalized analytic continuation of derivative
Observe that $$H'(z)=\sum_{k\in F}\frac{\sqrt{k}e^{\sqrt{k}z}}{(1-e^{\sqrt{k}z})^2}=\sum_{k\in F}\frac{\sqrt{k}}{4}\cdot\text{sech}(\frac{\sqrt{k}z}{2})^2.$$ This formula converges on $\{a+bi\mid a,b\in\mathbb{R},a\neq 0\}$, so the derivative $H'$ admits a 'generalized analytic continuation' to the complex plane except for the imaginary axis. We have $H'(z)=H'(-z)$ whenever $z$ is a complex number that is not on the imaginary axis. However, I do not know of any rigorous generalized notion of analytic continuation in which we can extend $H'$ from the half plane so that it satisfies $H'(z)=H'(-z)$, and I cannot find out
Continuation to compactification
Now, let $\iota:\mathbb{R}\rightarrow(S^1)^F$ be the function defined by letting
$\iota(\theta)=(e^{i\sqrt{k}\theta})_{k\in F}$. Then $\iota[\mathbb{R}]$ is dense in $(S^1)^F$.
Define a function $K:(-\infty,0)\times(S^1)^F\rightarrow\mathbb{C}$ by letting
$K(t,(b_k)_{k\in K})=\sum_{k\in F}\frac{b_k \exp(\sqrt{k}t)}{1-b_k\exp(\sqrt{k}t)}.$
Then the function $K$ is an extension of the function $H$ since
$H(a+bi)=K(a,(e^{i\sqrt{k}b})_{k\in F})$. Then the function $K$ is a continuous function, and $K$ is the only continuous function where $H(a+bi)=K(a,(e^{i\sqrt{k}b})_{k\in F})$ whenever $a<0$. If $(t,(b_k)_{k\in K})\in(-\infty,0)\times S_1^F$, then
define a chart $j_{(t,(b_k)_{k\in K})}:\{z\in\mathbb{C}\mid\text{Re}(z)<0\}\rightarrow (-\infty,0)\times S_1^F$ by letting
$j_{(t,(b_k)_{k\in K})}(a+bi)=(a+t,(b_k\cdot e^{i\sqrt{k}b})_{k\in K}).$ Then the composition $K\circ j_{(t,(b_k)_{k\in F})}$ is always holomorphic. Therefore, $K$ is 'holomorphic' in the sense that the composition of $K$ with each of these coordinate charts is holomorphic and the images of these coordinate charts partition the space $(-\infty,0)\times S_1^F$ into a whole bunch of half planes.
Observe that $(-\infty,0)\times S_1^F$ is a locally compact abelian semigroup. We shall write $+$ for this semigroup operation.
Now, observe that $$\frac{nz^{n}}{z^{n}-1}=\sum_{k=0}^{n-1}\frac{z}{z-\exp(\frac{2\pi i k}{n})}.$$
Theorem: $$\sum_{j=0}^{n-1}K(\mathbf{z}+(e^{2\pi ij/n})_{k\in F})=n\cdot K(n\mathbf{z})$$ whenever $\mathbf{z}\in (-\infty,0)\times S_1^F.$
Proof: We first observe that $$-K(t,(e^{i\theta}b_k)_{k\in F})=\sum_{k\in F}\frac{b_k\exp(\sqrt{k}t)}{b_k\exp(\sqrt{k}t)-e^{i\theta}}.$$
Therefore,
$$\sum_{j=0}^{n-1}-K(t,(b_k\exp(-2\pi i j/n))_{k\in F})=\sum_{k\in F}\sum_{j=0}^{n-1}\frac{b_k\exp(\sqrt{k}t)}{b_k\exp(\sqrt{k}t)-\exp(2\pi ij/n)}$$
$$=\sum_{k\in F}\frac{n(b_k\exp(\sqrt{k}t))^n}{(b_k\exp(\sqrt{k}t))^n-1}
=n\sum_{k\in F}\frac{b_k^n\exp(n\sqrt{k}t)}{b_k^n\exp(n\sqrt{k}t)-1}
=-n\cdot K(nt,(b_k^n)_{k\in F}).$$ Q.E.D.
As a consequence, $$\sum_{j=0}^{n-1}K'(\mathbf{z}+(e^{2\pi ij/n})_{k\in F})=n^2\cdot K'(n\mathbf{z})$$ whenever $\mathbf{z}\in (-\infty,0)\times S_1^F.$
A: The series
$$\tag{0}\label{0}
G_a(x)=\sum_{n=0}^\infty x^{\sqrt[a]{n}}
$$
can be calculated for $0<x<1$ and $a\geq1$ using Cauchy's residue theorem. We use the fact that $\pi \cot(\pi n)$ has residues 1 at $n\in \mathbb Z$, and deform the integration contour to the imaginary axis in the usual way. After the substitution $n\to (-\nu\,/\log x)^a$ we get
\begin{align}\tag{1}\label{1}
G_a(x)
%&= 
%\frac{\Gamma(1+a)}{(-\log x)^a} + \frac 1 2
%+ \frac{2a}{(-\log x)^a} \times\\
%&\quad\int_0^\infty \nu^{a-1} 
%\exp\left[-\nu \cos\left(\frac{\pi}{2a}\right)\right] 
%\sin\left[ \nu \sin\left(\frac{\pi}{2a}\right)\right]
%\left(\coth\left[\pi \left(\frac{\nu}{-\log %x}\right)^a\right]-1\right)\mathrm{d}\nu \\
&=\frac{\Gamma(1+a)}{(-\log x)^a} + \frac 1 2 
+ \frac{2a}{(-\log x)^a} \int_0^\infty \frac{\nu^{a-1} 
\exp\left[-\nu \cos\left(\frac{\pi}{2a}\right)\right] 
\sin\left[ \nu \sin\left(\frac{\pi}{2a}\right)\right]}
{\exp\left[2\pi \left(\frac{\nu}{-\log x}\right)^a\right]-1} \mathrm{d}\nu,
\end{align}
or, after back-substitution,
$$\tag{2}\label{2}
G_a(x)= \frac{\Gamma(1+a)}{(-\log x)^a} + \frac 1 2
+ \int_0^{\mathrm i\infty} \frac{ 
x^{\sqrt[a]{n}} - x^{\sqrt[a]{-n}}}
{\mathrm e^{-2\pi \mathrm i n}-1} \mathrm{d} n.
$$
The $1/2$ can be seen as the first Euler–Maclaurin correction. The integral \eqref{2} converges exponentially fast. This result is identically obtained using the Abel–Plana formula.
Edit 08.08.22,23:20 CEST
As pointed out by @Joseph, we better consider the series
$$\tag{3}\label{3}
H_a(z) = G_a(\mathrm e^z)-1 = \sum_{n=1}^\infty \mathrm e^{\sqrt[a]{n} z},
$$
with $\mathrm{Re}(z) < 0$ and again $a\geq1$.
An even simpler evaluation is the following: expanding the exponential into a Taylor series,
$$\tag{4}\label{4}
\mathrm e^{\sqrt[a]{n} z} = 1 + \sum_{k=1}^\infty n^{k/a} \frac{z^k}{k!}
$$
and interchanging sums, we directly get a representation involving the Riemann zeta function,
$$\tag{5}\label{5}
H_a(z) = \frac{\Gamma(1+a)}{(-z)^{a}} + \sum_{k=0}^\infty \zeta\left(-\frac{k}{a}\right)\frac{z^k}{k!},
$$
where the first term is regularized as in \eqref{2}.
The term $-1/2$ is identified with the zeta-regularized sum for $k=0$ and is moved back into the sum, which therefore starts at $k=0$.
A Mathematica function that checks \eqref{3} against \eqref{5} reads
r[a_,z_] := Gamma[1+a]/(-z)^a + Sum[Zeta[-k/a] z^k/k!, {k,0,100}]
            - NSum[Exp[n^(1/a) z], {n,1,∞}, NSumTerms->200000, 
              WorkingPrecision->50, Method->"EulerMaclaurin"]

I guess that for the original case $a=2$, Eq. \eqref{5} can be related to square-free numbers from @Joseph's answer through their generating function $\zeta(s)/\zeta(2s)$, cf. https://en.wikipedia.org/wiki/Square-free_integer.
