TITLE: the ratio of jacobi theta functions and a new conjectured q-continued fraction
QUESTION [9 upvotes]: Given the squared nome $q=e^{2i\pi\tau}$ with $|q|\lt1$, define 
$$\begin{aligned}H(q)=\cfrac{2(1+q^2)}{1-q+\cfrac{(1+q)(1+q^3)}{1-q^3+\cfrac{2q^2(1+q^4)}{1-q^5+\cfrac{q^3(1+q)(1+q^5)}{1-q^7+\cfrac{q^4(1+q^2)(1+q^6)}{1-q^9+\cfrac{q^5(1+q^3)(1+q^7)}{1-q^{11}+\ddots}}}}}}\end{aligned}$$
Question: How do we show that $$H(q)\overset{\color{red}{?}}=\frac{2\,q^{1/2}\,\vartheta_3(0,q^2)}{\vartheta_2(0,q^2)}$$
where $\vartheta_n(0,q)$ are jacobi theta functions

REPLY [6 votes]: In a related answer,
I used a formula by Ramanujan, proved by Adiga et al. (1985):
$$\small\frac{(-a;q)_\infty\,(-b;q)_\infty - (a;q)_\infty\,(b;q)_\infty}
{(-a;q)_\infty\,(-b;q)_\infty + (a;q)_\infty\,(b;q)_\infty}
= \cfrac{a+b}{1-q+\cfrac{(a+bq)(aq+b)}{1-q^3+\cfrac{q\,(a+bq^2)(aq^2+b)}
{1-q^5+\cfrac{q^2(a+bq^3)(aq^3+b)}{1-q^7+\cdots}}}}\tag{*}$$
Applying it here can be done by setting $a=q_2^3$, $b=q_2^{-1}$
where $q_n=\exp(2\pi\mathrm{i}\tau/n)$, so $q_n^n=q$. Thus we get
$$\begin{align}
    \frac{H(q)}{2q_2} &=
    q_2^{-1}\,\cfrac{1+q^2}{1-q+\cfrac{(1+q)(1+q^3)}{1-q^3+\cfrac{2q^2(1+q^4)}
    {1-q^5+\cfrac{q^3(1+q)(1+q^5)}{1-q^7+\cfrac{q^4(1+q^2)(1+q^6)}
    {1-q^9+\cdots}}}}}
\\  &= \frac{(-q_2^3;q)_\infty\,(-q_2^{-1};q)_\infty
    - (q_2^3;q)_\infty\,(q_2^{-1};q)_\infty}
    {(-q_2^3;q)_\infty\,(-q_2^{-1};q)_\infty
    + (q_2^3;q)_\infty\,(q_2^{-1};q)_\infty}
\\  &= \frac{(-q_2^3;q)_\infty^2\,(-q_2^{-1};q)_2
    - (q_2^3;q)_\infty^2\,(q_2^{-1};q)_2}
    {(-q_2^3;q)_\infty^2\,\underbrace{(-q_2^{-1};q)_2}_{(1+q_2)^2/q_2}
    + (q_2^3;q)_\infty^2\,\underbrace{(q_2^{-1};q)_2}_{-(1-q_2)^2/q_2}}
\\  &= \frac{(-q_2;q)_\infty^2 + (q_2;q)_\infty^2}
    {(-q_2;q)_\infty^2 - (q_2;q)_\infty^2}
\\  &\stackrel{[1]}{=}
    \frac{\vartheta_3(0,q_2) + \vartheta_3(0,-q_2)}
    {\vartheta_3(0,q_2) - \vartheta_3(0,-q_2)}
\\  &\stackrel{[2]}{=}
    \frac{\vartheta_3(0,q^2)}{\vartheta_2(0,q^2)}
\end{align}$$
where [1] follows from the product representation
$$\begin{align}
    \vartheta_3(0,q_2) &= (-q_2;q)_\infty^2\,(q;q)_\infty
\end{align}$$
and [2] follows from splitting the series representation of $\vartheta_3$
into odd and even parts, resulting in:
$$\begin{align}
    \vartheta_3(0,q_2) + \vartheta_3(0,-q_2) &= 2\vartheta_3(0,q^2)
\\  \vartheta_3(0,q_2) - \vartheta_3(0,-q_2) &= 2\vartheta_2(0,q^2)
\end{align}$$
More details and background on the Ramanujan formula can be found
at the other answer.