Rogers–Ramanujan continued fraction

Remarks

Since $\abs{q} < 1$ the limit exists, and Rogers and Ramanujan showed $$R(\tau) = q^{1/5} \prod_{n=1}^{\infty} \frac{(1 - q^{5n-1})(1 - q^{5n-4})}{(1 - q^{5n-2})(1 - q^{5n-3})} = q^{1/5}\, \frac{H(q)}{G(q)},$$ with $G(q) = \sum_{n \ge 0} \frac{q^{n^2}}{(q; q)_n}$ and $H(q) = \sum_{n \ge 0} \frac{q^{n^2 + n}}{(q; q)_n}$, the two sides of the Rogers–Ramanujan identities.

Example: $\tau = i$, so $q = e^{-2\pi} \approx 0.001867$. Then $R(i) = \sqrt{\frac{5 + \sqrt5}{2}} - \frac{1 + \sqrt5}{2} \approx 0.284079$. A numerical check: $q^{1/5} = e^{-2\pi/5} \approx 0.284609$, and the continued fraction $1 + q/(1 + q^2/(1 + \cdots))$ is about $1.001867$, whose quotient is about $0.284079$.

Non-example: with numerators $1, 1, 1, \dots$ instead of $q, q^2, q^3, \dots$, the fraction $\cfrac{1}{1 + \cfrac{1}{1 + \cdots}}$ is $\frac{\sqrt5 - 1}{2}$, the inverse of the golden ratio, and it does not depend on $q$; it is not $R$.

Another common notation is $R(q)$ in place of $R(\tau)$.