# 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?

• Can you have a close look at your definition for $R_2$ ... $R_{2}=\frac{1}{\Big(\frac{1}{q^2}\Big(\frac{1}{\Big(\frac{1}{q}\Big(\frac{1}{R(q)}-1\Big)\Big)}\color{red}{-1}\Big))}$ – Donald Splutterwit Jul 22 '17 at 20:28
• If your goal is to calculate $R'(q)$, you might consider using the fact that $$R(q) = q^{1/5}\prod_{k=0}^{\infty} \frac{(1-q^{5k+1})(1-q^{5k+4})}{(1-q^{5k+2})(1-q^{5k+3})}$$ – Michael Lee Jul 22 '17 at 21:09
• @MichaelLee How did you get that? If I may ask. – Simply Beautiful Art Jul 22 '17 at 21:32
• L. J.. Rogers first and then S. Ramanujan later both proved it over a hundred years ago. See the Wikipedia article Rogers-Ramanujan identities. – Somos Jul 22 '17 at 22:19
• Note that $\frac{1}{R_n} = R(q^n, q)$ where $R(a,q)$ is the generalized Rogers-Ramanujan continued fraction mentioned at Mathworld (37). It is often easier to prove statements about $R(a,q)$ because of the recurrence relation $R(a,q)=\frac{1}{1+aq\,R(aq,q)}$. – ccorn Jul 23 '17 at 7:55

A little study of $\,R_n\,$ turns up the result $\,R_n = 1 + q^{n+1} / R_{n+1}\,$ for $n\ge 0$. Taking a derivative gives $$R_n' = \frac{(n+1)q^n R_{n+1} - q^{n+1}R_{n+1}'}{R_{n+1}^2}$$ 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 our result as $$Q_n = (n+1)q^nP_n - q^{n+1}P_nQ_{n+1}.$$ By iteration we get $$Q_0 = 1q^0P_0 - q^1P_0Q_1 = 1q^1P_0 -q^1P_0(2q^1P_1 - q^2P_1Q_2)$$ and taking the limit $$Q_0 = 1q^0P_0 - 2q^2P_0P_1 + 3q^5P_0P_1P_2 - 4q^9P_0P_1P_2P_3 + \dots$$ which is our 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 $\, R_n = F(q^{n+1},q). \,$
• @Nicco Thanks, I like it myself for getting it. You may want to fix the sign error in your equation (1) since, as I mention, the left side should be $-R'(q)/R(q)$. – Somos Jul 23 '17 at 17:43