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.



Is anything known about the series $\sum_{n=0}^{\infty} x^{\sqrt{n}} $?
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](https://en.wikipedia.org/wiki/Abel%E2%80%93Plana_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>.


