In his famous paper "Modular Equations and Approximations to $\pi$", Ramanujan gives the following famous series for $1/\pi$:
\begin{align}\frac{1}{2\pi\sqrt{2}} &= \frac{1103}{99^{2}} +
\frac{27493}{99^{6}}\frac{1}{2}\frac{1\cdot 3}{4^{2}} +
\frac{53883}{99^{10}}\frac{1\cdot 3}{2\cdot 4}\frac{1\cdot 3\cdot 5\cdot
7}{4^{2}\cdot 8^{2}} + \cdots\notag\\
&= \sum_{n = 0}^{\infty}\dfrac{\left(\dfrac{1}{4}\right)_{n}\left(\dfrac{1}{2}\right)_{n}\left(\dfrac{3}{4}\right)_{n}}{(n!)^{3}}(1103 + 26390n)\left(\frac{1}{99^{2}}\right)^{2n + 1}\notag\end{align}
He also mentions the technique for finding such series which is based on the evaluation of $nP(q^{n}) – P(q)$ in a closed form. Here $$P(q) = 1 – 24\sum_{j = 1}^{\infty}\frac{jq^{2j}}{1 – q^{2j}}$$ The kind of closed form needed is $$nP(q^{n}) – P(q) = \frac{4LK}{\pi^{2}}\cdot A(l, k)$$ where $k, l$ and $K, L$ correspond to $q, q^{n}$ and $A(l, k)$ is an algebraic function. In order to derive the series mentioned above it is necessary to calculate this expression $nP(q^{n}) – P(q)$ for $n = 58 = 2\cdot 29$ which can be done (as mentioned by Ramanujan) if we can calculate its value for $n = 2$ and $n = 29$. Sadly Ramanujan does not give the expression $A(l, k)$ for $n = 29$. I consulted books of Bruce C. Berndt but could not find this specific expression. Although Ramanujan mentions a process where this expression can be obtained from a modular equation of degree $29$, but due to the complexity of Russell's modular equation of degree $29$ I can't apply the technique.
Is there any work (paper) available which tries to directly use Ramanujan's approach and prove the above series by calculating $A(l, k)$ for $n = 29$ from a modular equation?
Note: There does not seem to be (Edit: this has changed since then) a specific tag related to Ramanujan so I have put this under "sequences-and-series" and noting that nowadays most of Ramanujan's work is studied under modular-forms I have added that tag.
Best Answer
I would like to outline a proof of this famous identity, which is closely related to the question I have posted on MathOverflow. It is somewhat different from the proof of Borwein brothers.
Addendum: I can use a construction given by Mazur-Swinnerton-Dyer and Zagier to prove Ramanujan's identity for $n=37$.
Let $n=58$.
1. Following Borwein brothers, we can get
$$\frac{1}{\pi}=\sum_{m=0}^{\infty}(2\sqrt{n}v(k)m+G_0)b_mc^m(k)$$
where
$$b_m=\frac{(4m)!}{4^{4m}(m!)^4}$$
$$2v(k)=\left(1-\frac{2}{((k^{\prime})^2/(2k))^2+1}\right)$$
$$c(k)=\left(\frac{2}{2k/(k^{\prime})^2+(k^{\prime})^2/(2k)}\right)^2$$
$$G_0=\frac{\sqrt{n}}{3}\left(1-\frac{3}{2(((k^{\prime})^2/(2k))^2+1)}-\frac{1}{1+k^2}\frac{G_1}{2}\right)$$
$$G_1=\frac{nP(q^n)-P(q)}{(2K(k)/\pi)^2}$$
2. Following H. M. Weber, one has
$$\frac{2k}{(k^{\prime})^2}=\left(\frac{\sqrt{29}-5}{2}\right)^6$$
where $k=k(e^{-\pi\sqrt{58}})$
3. $$\frac{G_1}{1+k^2}=\frac{nP(q^n)-P(q)}{\eta^{2}(q^{2n})\eta^{2}(q^{4})}\frac{c(k)^{1/4}}{8\sqrt{n}}$$
Denote $$H(q)=\frac{nP(q^n)-P(q)}{\eta^{2}(q^{2n})\eta^{2}(q^{4})}$$
Then $H(q)^2$ is a weakly modular form on $\mathbb{H}/\Gamma_0(58)$
4. Denote $$[a_1,\cdots,a_n]=\prod_{\delta\mid N,\sum a_\delta=0}\eta^{a_\delta}(\delta\tau)$$
where $\eta$ is Dedekind eta function. Then $H(q)^2\cdot[-2,8,10,-16]$ is holomorphic on $\mathbb{H}/\Gamma_0(58)$ except at infinity. Then it is a linear combination of eta function product invariant under $\Gamma_0(58)$.
5. $[\cdots,\alpha_\delta,\cdots]$ is invariant under $\Gamma_0(58)$ if 1)$24\mid\sum_{\delta\mid 58}\delta a_{\delta}$; 2) $24\mid\sum_{\delta\mid 58}58a_{\delta}/\delta $ ; 3)$\prod_{\delta \mid n}\delta^{a_\delta}$ is a rational square. What's more, eta product is holomorphic at a cusp $c/d$ if
$$\frac{1}{24}\sum_{\delta\mid 58}\frac{(\mathrm{gcd}(d,\delta))^2}{\delta}a_\delta\geq 0$$
6. Expand $H(q)^2\cdot[-2,8,10,-16]$ and eta products, use matkerint function in PARI/GP to calculate the coefficients of linear combination. Note that
$$[a,b,c,d]=2^{c/2}58^{(a+b)/4}\left(\frac{\sqrt{2}}{2}\frac{\sqrt{29}+5}{2}\right)^{(a+d)/2}$$
where $q=\exp(-\pi/\sqrt{58})$. All these would lead to
$$G_0=\frac{\sqrt{58}}{3}\left(1-\frac{3}{4\times 99^2}\left(\frac{\sqrt{29}-5}{2}\right)^6-\frac{36\sqrt{2}(148 + 11 \sqrt{29})}{99\times16\sqrt{58}}\right)$$
and we are done.