Ramanujan's ability to prove theorem's and define new functions was remarkable. In his very short life (1887-1920), he manage to rediscover all elliptic and theta functions theory that take centuries to developed. His work is in research until nowdays. He did not leave us writing proof's, because he has no time to do this. For example in [BerIII] one can see that firstly he defined the basic ''null'' theta functions and then construct from them a huge number of theorem's, from the very simple ones, to modular equations of very high degree say 255. This may seen as: If \begin{equation}
K(x)=\int^{\pi/2}_{0}\frac{d\theta}{\sqrt{1-x^2\sin^2(\theta)}}=\frac{\pi}{2}{}_2F_1\left(\frac{1}{2},\frac{1}{2};1;x^2\right),
\end{equation}
is the complete elliptic integral of the first kind. The elliptic singular modulus $k_r$ is defined from the equation (I use the traditional notation)
$$
\frac{K\left(\sqrt{1-x^2}\right)}{K\left(x\right)}=\sqrt{r}\Leftrightarrow x=k_r\in(0,1)
$$
and
$$
z_n:={}_2F_1\left(\frac{1}{2},\frac{1}{2};1;k^2_{n^2r}\right)=K(k_{n^2r})=K[n^2r]\textrm{, }n=1,2,3,\ldots,
$$
then what is
$$
m_n=m_n(r)=\frac{z_n}{z_1}=\frac{K[n^2 r]}{K[r]}=?
$$
(In general if $r$ is a positive rational, then both $k_r$and $m_n(r)$ are algebraic numbers).
In his notes, Hardy commented that Ramanujan had produced groundbreaking new theorems, including some that "defeated me completely; I had never seen anything in the least like them before".
Returning to Berndt, in his book says (litle rephrased by me): ''We don't have proof's. Our proof's are perhaps more aptly may called verifications''. He then uses computer program MACSYMA to verify "prove" these results, using not the theory of tranformation of theta functions and $q-$series, but rather results from the theory of modular forms developed much later. Berndt uses as general references books of Rankin in 1977 and Pettersson 1974 (see [BerIII] pg. 326 and its related references).
He also had no book's to read. He did not pass in the exams in the university (because he was bad at English). The Rogers-Ramanujan continued fraction and the specific evaluation you mention, was only a very-very small touch οn a very-very large board. While many great mathematicians of the 20th century were enjoying reading papers of Euler, Gauss, Dirichlet, Jacobi, Eisenstein, Weber, Kronecker, Hurwitz, Riemann and others, Ramanujan open this way alone and passed them.
About the Rogers-Ramanujan Continued Fraction
Ramanujan have first discovered that:
For $|q|<1$ and $|z|<|q|^{-1}$ a more general version of Rogers Ramanujan continued fraction is
\begin{equation}
R(z;q):=\frac{q^{1/5}}{1+}\frac{qz}{1+}\frac{q^2z}{1+}\frac{q^3z}{1+}\ldots
\end{equation}
For $|q|<1$ we also define
\begin{equation}
(a;q)_n:=\prod^{n-1}_{k=0}{(1-aq^k)}\textrm{, } (a;q)_0:=1
\end{equation}
and
\begin{equation}
(q)_n:=(q;q)_n\textrm{, }(q)_0:=1.
\end{equation}
Set also
$$
f(-q):=(q;q)_{\infty}
$$
Theorem 1.(See [Berndt] chaper 16, Entry 15)
If $|q|<1$ and
\begin{equation}
H(z;q):=\sum^{\infty}_{n=0}\frac{q^{n(n+1)}z^n}{(q)_n}\textrm{, }
G(z;q):=\sum^{\infty}_{n=0}\frac{q^{n^2}z^n}{(q)_n},
\end{equation}
then
\begin{equation}
R(z;q)=q^{1/5}\frac{H(z;q)}{G(z;q)}.
\end{equation}
Having in mind the above he proceeded and find
Theorem 2. (The Rogers-Ramanujan identities) (see [Andrews]: Sec. 14-3,14-4)
If $|q|<1$, then
\begin{equation}
H(1;q)=f(-q)^{-1}\sum^{\infty}_{n=-\infty}(-1)^nq^{5n^2/2+3n/2}=\prod^{\infty}_{n=0}\frac{1}{(1-q^{5n+2})(1-q^{5n+3})}
\end{equation}
and
\begin{equation}
G(1;q)=f(-q)^{-1}\sum^{\infty}_{n=-\infty}(-1)^nq^{5n^2/2+n/2}=\prod^{\infty}_{n=0}\frac{1}{(1-q^{5n+1})(1-q^{5n+4})}.
\end{equation}
Theorem 3. For all $|q|<1$, we have (we denote $R(q):=R(1;q))$:
\begin{equation}
\frac{1}{R(q)}-1-R(q)=\frac{f(-q^{1/5})}{q^{1/5}f(-q^5)}
\end{equation}
and
\begin{equation}
\frac{1}{R(q)^5}-11-R(q)^5=\frac{f^6(-q)}{q f^6(-q^5)}.
\end{equation}
Theorem 4. (see [BerIII] Chapter 16, Entry 39). If $a,b>0$ and $ab=\pi^2$, then
\begin{equation}
\left\{\frac{\sqrt{5}+1}{2}+R(e^{-2a})\right\}\cdot\left\{\frac{\sqrt{5}+1}{2}+R(e^{-2b})\right\}=\frac{5+\sqrt{5}}{2}.
\end{equation}
Remarks. From Jacobi's theory of elliptic functions [W,W] pg. 488 we have the next exersize
$$
f(-q^2)^6=2\pi^{-3}q^{-1/2}k_rk'_rK^3\textrm{, this was known to Ramanujan,}
$$
where $q=e^{-\pi\sqrt{r}}$, $r>0$, $k'_r:=\sqrt{1-k_r^2}$.
Also the lower modular equations and its multipliers are
$$
K=K(k_r)=z_1\textrm{, }m_2=\frac{z_2}{z_1}=\frac{1+k'_r}{2}\textrm{, }k_1=\frac{1}{\sqrt{2}},
$$
$$
k_{4r}=\frac{1-k'_r}{1+k'_r}\textrm{, the modular equation of degree 2.}
$$
(Multipliers of degree 3 and 5):
\begin{equation}
27m^4_3-18m^2_3-8(1-2k^2_r)m_3-1=0
\end{equation}
\begin{equation}
(5m_5-1)^5(1-m_5)=256k^2_r(1-k^2_r)m_5
\end{equation}
As an example Ramanujan have calculated (see [Vil])
$$
k_{210}^2=(4-\sqrt{15})^4(8-3\sqrt{7})^2(2-\sqrt{3})^2(6-\sqrt{35})^2(\sqrt{10}-3)^4 (\sqrt{7}-\sqrt{6})^4\times
$$
$$
\times(\sqrt{15}-\sqrt{14})^2(\sqrt{2}-1)^2.
$$
REFERENCES
[Andrews]: G.E. Andrews, ''Number Theory''. Dover Publications, New York. 1994.
[BerIII]: B.C.Berndt, ''Ramanujan`s Notebooks Part III''. Springer Verlag, New York (1991)
[Vil]: Mark B. Villarino. ''Ramanujan's Most Singular Modulus''. arXiv:math/0308028v4 [math.HO] 17 Jul 2005.
[W,W]: E.T.Whittaker and G.N.Watson, ''A course on Modern Analysis''. Cambridge U.P. (1927)
