Evaluation of $\sum_{n=1}^{\infty}q^{n^{2}}$?

Question

I would like to evaluate the following summation \begin{equation} \sum_{n=1}^{\infty}q^{n^{2}} \end{equation} assuming $ 0<q<1$ obviously the series converge but can anyone help me how to evaluate it? I know it's related to Jacobi theta function fro some special value but that is not what I'm looking for as an answer.

Answer 1

I think you are almost there: the Jacobi's third theta function in nome form is $$ \vartheta_3(\eta, q) = \sum_{n=-\infty}^{\infty} q^{n^2} \eta^n. $$ Thus your series can be evaluated via $\frac{1}{2}(\vartheta_3(1,q)-1)$. Most scientific computing packages allow you to do so. For example, below is the plot of the series against $0<q<1$ using the mpmath package (note $\vartheta_3(1,q)$ corresponds to mpmath.jtheta(n=3, z=0, q=q).)

$0<q<1$, $q\in\mathbb{R}$">

EDIT: Minimax Rational Approximation

We will write $\vartheta_3(q)$ for $\vartheta_3(1,q)$. As OP asked for analytical approximation when $q\in\mathbb{R}$ with small values, it might be possible to consider as simple as a rational function of the form $$ r(q) = \frac{\sum_{i=0}^n a_i q^i}{1 + \sum_{i=1}^m b_i q^i}. $$ Note that $r(q)$ is quasi-convex, and thus we can generate a bunch of evaluations $\{(q_i, y_i)\}_{i=1}^N$ with $y_i=\vartheta_3(q_i)$ and do a minimax rational fit (c.f. Boyd Ex. A 5.2) to minimize the max of $\lvert{r(q_i) - \vartheta_i(q_i)}\rvert$. For example, using $N=1000$ and $n=m=3$ we obtained a fit $$ \hat{\mathbf{a}} = {[ 1.044359498, \quad -1.271249380, \quad -0.44275578, \quad 0.670133658 ]}^{\top}, \quad \hat{\mathbf{b}} = {[ -2.56134660,\quad 2.129914536,\quad -0.568564211 ]}^{\top} $$ The maximum absolute error is below $0.05$ for $q$ bounded below from $1$, and well below $0.05$ for most of the values (see below). It should be feasible to study bounds on the derivate of $\vartheta_3(q)$ with small values of $q$, hence the error bound can be made uniform in that region. And certainly you can use more samples and finer stopping criteria to meet your accuracy need.

$m=n=3$.">