Does the Rogers-Ramanujan continued fraction $R(q)$ satisfy this conjectured infinite series

Given the Rogers-Ramanujan continued fraction

$R(q)= \cfrac{1}{1+\cfrac{q}{1+\cfrac{q^2}{1+\cfrac{q^3}{1+\ddots}}}}$

where $q=\exp(2\pi i \tau)$, $|q|\lt1$

for the sake of brevity, let us introduce the following notation

$R_{0}=\frac{1}{R(q)}$

$R_{1}=\frac{1}{\Big(\frac{1}{q}\Big(\frac{1}{R(q)}-1\Big)\Big)}$

$R_{2}=\frac{1}{\Big(\frac{1}{q^2}\Big(\frac{1}{\Big(\frac{1}{q}\Big(\frac{1}{R(q)}-1\Big)\Big)}-1\Big)\Big)}$

$R_{3}=\frac{1}{\Big(\frac{1}{q^3}\Big(\frac{1}{\Big(\frac{1}{q^2}\Big(\frac{1}{\Big(\frac{1}{q}\Big(\frac{1}{R(q)}-1\Big)\Big)}-1\Big)\Big)}-1\Big)\Big)}$

up to $R_{n}$, $\frac{1}{R_{n}}=R(q^n,q)$ for natural number $n$, where $R(a,q)$ is the Generalized Rogers-Ramanujan continued fraction as pointed out by @ccorn

It is then conjectured that the following infinite series holds

$-\frac{R'(q)}{R(q)}=\frac{1}{R_{0}R_{1}}-\frac{2q^2}{R_{0}R^2_{1}R_{2}}+\frac{3q^5}{R_{0}R^2_{1}R^2_{2}R_{3}}-\frac{4q^9}{R_{0}R^2_{1}R^2_{2}R^2_{3}R_{4}}+\frac{5q^{14}}{R_{0}R^2_{1}R^2_{2}R^2_{3}R^2_{4}R_{5}}-\dots\tag1$

It has the ascending continued fraction equivalent

$-\frac{R'(q)}{R(q)}= \frac{\frac{1}{R_1}+\large{\frac{\frac{-2q^2}{R_2}+\large{\frac{\frac{3q^5}{R_3}+...}{R^2_2}}}{R^2_1}}}{R_0} \tag2$

After considering the Rogers-Ramanujan continued fraction with the factor $q^{1/5}$ and applying the identity $\frac{R'(q)}{R(q)}=\frac{1}{5q}\frac{(q;q)^5_{\infty}}{(q^5;q^5)_{\infty}}$ due to Ramanujan,we are led to the following beautiful identity

$\frac{(q)^5_{\infty}}{(q^5)_{\infty}}=1-\frac{5q}{(R_{0}R_{1})}+\frac{10q^3}{(R_{0}R_{1})(R_{1}R_{2})}-\frac{15q^6}{(R_{0}R_{1})(R_{1}R_{2})(R_{2}R_{3})}+\frac{20q^{10}}{(R_{0}R_{1})(R_{1}R_{2})(R_{2}R_{3})(R_{3}R_{4})}-\frac{25q^{15}}{(R_{0}R_{1})(R_{1}R_{2})(R_{2}R_{3})(R_{3}R_{4})(R_{4}R_{5})}+\dots\tag3$

$\frac{(q)^5_{\infty}}{(q^5)_{\infty}}=1+5\sum_{n=1}^{\infty}\frac{(-1)^n nq^{\frac{n(n+1)}{2}}}{\prod_{k=1}^{n-1}\Big(R_{k}+q^k\Big)}$

Question:How do we prove that the conjecture is true?

Solution 1:

A little study of $\,R_n\,$ turns up the result $\,R_n = 1 + q^{n+1} / R_{n+1}\,$ for $n\ge 0$.

Take the derivative to get $$R_n^{\,'} = \frac{(n+1)q^n R_{n+1} - q^{n+1}R_{n+1}^{\,'}}{R_{n+1}^2}$$ and then $$\frac{R_n^{\,'}}{R_n} = \frac{ (n+1)q^n }{(R_nR_{n+1})} - \frac{q^{n+1}}{(R_nR_{n+1})}\frac{R_{n+1}^{\,'}}{R_{n+1}}.$$

For brevity, define $\, P_n := \frac{1}{R_nR_{n+1}},\: Q_n := \frac{R_n^{\,'}}{R_n} \,$ and rewrite the result as $$ Q_n = (n+1)q^nP_n - q^{n+1}P_nQ_{n+1}.$$ Use iteration to get $$ Q_0 = 1q^0P_0 - q^1P_0Q_1 = 1q^1P_0 -q^1P_0(2q^1P_1 - q^2P_1Q_2) $$ and take the limit to get $$ Q_0 = 1q^0P_0 - 2q^2P_0P_1 + 3q^5P_0P_1P_2 - 4q^9P_0P_1P_2P_3 + \dots $$ which is the result, but notice, since $\, R(q) := 1/R_0, \,$ the left side is also $\, -R^{\,'}\!(q)/R(q). \,$

By the way, if we define $\, F(x,q) = 1 + x/F(xq,q),\,$ then $\, R_n = F(q^{n+1},q). \,$