Write $\sum\limits_{n=0}^\infty e^{-xn^3}$ in the form $\sum\limits_{n=-\infty}^\infty a_nx^n$ This is a very simple question; I apologize if it has already been asked here. Define the following function (superficially similar to a theta function):
$$\varsigma(x)=\sum_{n=1}^\infty e^{-xn^3}$$
I am interested in knowing the Laurent series about $x=0$ of this series if it exists, i.e. I would like to know if there exist $\{a_n\}$ such that:
$$\varsigma(x)=\sum_{n=-\infty}^\infty a_nx^n\tag{1}$$
for small enough $x>0$. I'm pretty sure that $\varsigma(x)$ diverges to infinity at $x=0$ so I assume that some of the $a_n$ for $n<0$ will be nonzero. Ideally I would love a closed form for the $a_n$ but I am especially interested for the moment in $a_1$. I have no idea how to find these terms, since this is not a Taylor series and since I do not know the complex behaviour of $\varsigma(z)$. Wolfram Alpha doesn't help me either. I am aware of the formula for Laurent series coefficients, and that for instance we will have:
$$a_1=\frac{1}{2\pi i}\oint_C \frac{\varsigma(z)}{z^2}\;dz$$
where $C$ is a closed contour around $z=0$, but I am not sure how I should go about evaluating this; formally interchanging integral and sum only gives me a divergent sum: for instance making use of the fact that $e^{az}=\sum\limits_{n=0}^\infty \frac{a^nz^n}{n!}$ I formally get $a_1=-\sum\limits_{n=1}^\infty n^3$ but I don't know whether I can make this argument rigorous to get $a_1=-\zeta(-3)$.
Background: This question arose from some recreational thoughts of mine on summing divergent series; this answer used an $ne^{-\epsilon n}$ regularization rather than the usual $n^s$ to 'evaluate' $\sum\limits_{n=1}^\infty n$ and curiously obtained a constant term of $-\frac{1}{12}$ in the Laurent series in $\epsilon$; this interested me and made me wonder what an $n^3e^{-\epsilon n^3}$ regularization of $\sum\limits_{n=1}^\infty n^3$ would give. We have $\sum\limits_{n=1}^\infty n^3e^{-\epsilon n^3}=-\varsigma'(\epsilon)$ so the constant term in the Laurent series expansion of this function will be $-a_1$; thus I would conjecture that $a_1=-\frac{1}{120}$ (i.e. $-\zeta(-3)$; see here). The above calculation supports this, but I don't know whether it can be made rigorous.
Thus I have the following questions: Do there exist $a_n$ such that $(1)$ holds for small $x>0$? If so, does $a_1=-\frac{1}{120}$? Is it possible to write a closed form for the $a_n$?
 A: In order for $\varsigma(x)$ to have a Laurent series in a deleted neighbourhood of $0$, $f$ must be analytic in that deleted neighbourhood.  But I'm pretty sure the function $\sum_n z^{n^3} = \varsigma(-\log(z))$ has a natural boundary on $|z|=1$.  Therefore no such Laurent series is possible.
EDIT:  The Fabry gap theorem says that if $\sum_n \alpha_n z^{p_n}$ has radius of convergence $1$, where $p_n$ is an increasing sequence of integers with $p_n/n \to \infty$, then the unit circle is a natural boundary for this series.
A: Very roughly, non rigorous
remove the n=0
$$ 1+\sum_{n=1}^{\infty}e^{-xn^3}$$
$$\sum_{n=1}^{\infty}e^{-xn^3}= \sum_{n=1}^{\infty}\sum_{m\in \mathbb{Z}} {(-xn^3)^m}{m!}$$
$$=\sum_{m\in \mathbb{Z}}e^{i\pi m/3}\frac{x^{m/3}\zeta(-m)\sum_{k=0}^2 (1/3)e^{\frac{2i\pi*km}{3}}}{(m/3)!}$$
If x is negative it needs to be in the same (complex) period as the alternating complex part. If any of periodical zero's of the function tend to grow in the in the neighbourhood of the zero's, either on minus inf or positive inf, rewrite the function with the reflection formula (for the zeta function), and by rewriting it as limits as the m goes tends to "a" 0. Then you can rewrite the zeta function back as sum again, to get some "hidden" sum.
Bit ashamed of a half answer, but it works also for most "theta" functions, and given it works for almost all sums, someone with more knowledge might be able to tell why and clean it up :)
Long story short, It's very important to define also the "zero" coefficients of the polynomal, because if you do not, when switching to the zeta function, it's just no the same sum.

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*one of the many problems with such a derivation, is that the complex root got multiple solutions, so again, this answer is horribly wrong, but it works sometimes and I will improve the answer later on.

