Conjectured continued fraction for the Jacobi theta function $\vartheta_{4}(q)$

Question

Given the Jacobi theta function $$\vartheta_{4}(q)=1+2\sum_{n=1}^{\infty}(-1)^nq^{n^2}$$

where $q=e^{2\pi i\tau}$, and $|q|\lt1$. It is conjectured that it has the following continued fraction

$\vartheta_{4}(q)= 1-\cfrac{2q}{1-q+\cfrac{q}{1-\cfrac{q^2}{1-q^3+\cfrac{q^2}{1-\cfrac{q^3}{1-q^5+\cfrac{q^3}{1-\cfrac{q^{4}}{1-q^7+\cfrac{q^{4}}{1-\dots}}}}}}}}\tag{1a}$

and its reciprocal,

$\begin{aligned}\frac{1}{\vartheta_{4}(q)}=1+\cfrac{2q}{1-q-\cfrac{q}{1+\cfrac{q^2}{1-q^3-\cfrac{q^2}{1+\cfrac{q^3}{1-q^5-\cfrac{q^3}{1+\cfrac{q^{4}}{1-q^7-\cfrac{q^{4}}{1+\dots}}}}}}}}\end{aligned}\tag{1b}$

How do we prove the conjecture?

Edited later:

If we define the q-pochhammer symbol for any complex number $a$, $$ (a;q)_{\infty}=\prod^{\infty}_{n=0}\left(1-aq^n\right) $$

the conjectured continued fractions can be generalized to the following form

$\begin{aligned}\prod_{n=0}^\infty\frac{\big(1-aq^n\big)}{\big(1+aq^n\big)}=1-\cfrac{2a}{1-q+\cfrac{a}{1-\cfrac{aq}{1-q^3+\cfrac{aq}{1-\cfrac{aq^2}{1-q^5+\cfrac{aq^2}{1-\cfrac{aq^{3}}{1-q^7+\cfrac{aq^{3}}{1-\dots}}}}}}}}\end{aligned}\tag{1c}$

which obviously leads to the continued fraction for the reciprocal of $\vartheta_{4}(q)$ when $a\rightarrow -q$.

Answer 1

This is a partial answer because I leave aside convergence considerations and just focus on the part I find interesting. It only treats the continued fraction labelled $(1\mathrm{a})$ in the question. The other ones have been added later.

Let $F(q)$ denote your continued fraction. Then $$\frac{1 + F(q)}{2} = b_0 + \dfrac{a_1}{b_1 + \dfrac{a_2}{b_2 + \ddots}}\tag{1}$$ where $$\begin{align} b_0 &= 1 & a_{2k-1} &= -q^k & a_{2k} &= q^k \\ & & b_{2k-1} &= 1 - q^{2k-1} & b_{2k} &= 1 \end{align}$$ for integer $k\geq 1$.

Here, I will focus on the even part of the continued fraction in $(1)$, that is, on the continued fraction whose sequence of approximants is the even-indexed subsequence of the approximants of $(1)$.

According to theorem 2.10 in reference 1 below, the even part is $$E(q) = b_0^* + \dfrac{a_1^*}{b_1^* + \dfrac{a_2^*}{b_2^* + \ddots}}$$ where $$\begin{align} b_0^* &= b_0 = 1 & a_1^* &= a_1 b_2 = -q \\ b_1^* &= a_2 + b_1 b_2 = 1 & a_2^* &= -a_2 a_3 b_4 = q^3 \\ b_k^* &= a_{2k-1}b_{2k} + b_{2k-2}(a_{2k} + b_{2k-1}b_{2k}) = 1 - q^{2k-1} & a_k^* &= -a_{2k-2}a_{2k-1}b_{2k-4}b_{2k} = q^{2k-1} \\ (k&\geq2) & (k&\geq 3) \end{align}$$ Thus $$E(q) = 1 - \dfrac{q}{1 + \dfrac{q^3}{1 - q^3 + \dfrac{q^5}{1 - q^5 + \ddots}}}$$ We recognize an Euler continued fraction therein: $$\dfrac{1}{1 + \dfrac{t_1}{1 - t_1 + \dfrac{t_2}{1 - t_2 + \ddots}}} = \sum_{n=0}^\infty(-1)^n\prod_{k=1}^n t_k$$ With $t_k = q^{2k+1}$ we get $$E(q) = 1 - q\sum_{n=0}^\infty(-1)^n\prod_{k=1}^n q^{2k+1} = \sum_{n=0}^\infty(-1)^n q^{n^2} = \frac{1 + \theta_4(q)}{2}$$ which confirms the claim for the even part of $(1)$ and offers the more compact $$\theta_4(q) = 2E(q) - 1 = 1 - \dfrac{2q}{1 + \dfrac{q^3}{1 - q^3 + \dfrac{q^5}{1 - q^5 + \ddots}}}$$ References:

1. W. B. Jones and W. J. Thron: Continued Fractions. Analytic Theory and Applications. Addison-Wesley 1980.