are these two continued fractions equivalent? I would like to pose the following conjecture.Given
$$\phi(q) =\cfrac{1}{1-q+\cfrac{q(1-q)^2}{1-q^3+\cfrac{q^3(1-q^2)^2}{1-q^5+\cfrac{q^5(1-q^3)^2}{1-q^7+\ddots}}}}$$ 
and
$$\psi(q)=\cfrac{-q}{1-q+\cfrac{q(1-q)^2}{1-q^3+\cfrac{q(1-q^2)^2}{1-q^5+\cfrac{q(1-q^3)^2}{1-q^7+\ddots}}}}$$ 
Then prove/disprove that the two continued fractions are equivalent
$$\phi(q)=\psi\left(\frac{1}{q}\right)$$
 A: So with formal manipulation:
$$\psi(1/q)= \frac{-q^{-1}}{1-q^{-1}+ \frac{q^{-1}(1-q^{-1})^2}{1-q^{-3}+\ldots}}$$
Times top and bottom by $-q$ to give:
$$\psi(1/q)= \frac{1}{1-q+ \frac{-qq^{-3}(1-q)^2}{1-q^{-3}+\frac{q^{-1}(1-q^{-2})^2}{1-q^{-5}+\ldots}}}$$
Then mutiply top and bottom of the next fraction by $-q^3$ to give:
$$\psi(1/q)= \frac{1}{1-q+ \frac{q(1-q)^2}{1-q^{3}+\frac{-q^3q^{-5}(1-q^{2})^2}{1-q^{-5}+\ldots}}}$$
Then times $-q^5$ gives 
$$\psi(1/q)= \frac{1}{1-q+ \frac{q(1-q)^2}{1-q^{3}+\frac{q^3(1-q^{2})^2}{1-q^{5}+\frac{-q^5q^{-7}(1-q^3)^2}{1-q^{-7}+\ldots}}}}$$
So letting the first denominator be line $1$, then in line $n$ we assume we carry a factor of $-q^{2n-1}$ which corrects the first term. In the second term, ie the top of the new denominator which is line $n+1$, we pull out a factor of $q^{-n}$ which times the $1\over q$ becomes $q^{-2n-1}$. Then carry this factor with the minus sign, through to the next line.
I imagine however this is actually why you assumed it was true in the first place and are looking more for a proof where you use series and convergence properties of generalised continued fractions. But from the formal algebraic standpoint above it seems true.
