the ratio of jacobi theta functions and a new conjectured q-continued fraction 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 
 A: 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.
A: (A partial answer.) Compare the three cfracs of similar form,
$$H(q)=\frac{q^{1/2}\,\vartheta_3(0,q^2)}{\vartheta_2(0,q^2)}=\small\cfrac{(1+q^2)}{1-q+\cfrac{q(1+q^{-1})(1+q^3)}{1-q^3+\cfrac{q^2(1+q^0)(1+q^4)}{1-q^5+\cfrac{q^3(1+q)(1+q^5)}{1-q^7+\cfrac{q^4(1+q^2)(1+q^6)}{1-q^9+\ddots}}}}}\tag1$$
$$U(-q)\; =\; 1\;=\;\small\cfrac{(1+q^1)}{1-q+\cfrac{q(1+q^0)(1+q^2)}{1-q^3+\cfrac{q^2(1+q)(1+q^3)}{1-q^5+\cfrac{q^3(1+q^2)(1+q^4)}{1-q^7+\cfrac{q^4(1+q^3)(1+q^5)}{1-q^{9}+\ddots}}}}}\tag2$$
$$S(q) = \frac{1}{q^{1/2}}\frac{\vartheta_2(0,q^2)}{\vartheta_3(0,q^2)}=\small\cfrac{(1+q^0)}{1-q+\cfrac{q(1+q)(1+q)}{1-q^3+\cfrac{q^2(1+q^2)(1+q^2)} {1-q^5+\cfrac{q^3(1+q^3)(1+q^3)}{1-q^7+\cfrac{q^4(1+q^4)(1+q^4)} {1-q^9+\ddots}}}}}\tag3$$
where the pattern of one cfrac to the next is clear. We have $(1)$ as your proposed equality, $(2)$ is the one in this post (with the small change $q \to -q$ for aesthetics), and $(3)$ was established by Somos in A079006. One can then see the nice fact that, 
$$H(q)\,U(-q)\,S(q) = 1$$
In fact, $H(q)$ and $S(q)$ are reciprocals. (This is the second reciprocal pair you have found after this, so I assume you are using a general method?)
P.S. Since $(2)$ belongs to an infinite family involving ratios of form $(1-aq^n)$, then $(1),(3)$ may have an infinite family as well. 
