I don't know much about Mathematica and its working so can't comment on its failure to find a closed form for the Ramanujan type series in your question.
It is worth noting that the series is not one of those listed in Ramanujan's 1914 paper Modular equations and approximations to $\pi$. However Ramanujan's technique can be applied to evaluate the sum of series in closed form.
To that end let $k\in(0,1)$ be the elliptic modulus, $k'=\sqrt{1-k^2}$ be the complementary modulus and let $$K(k) =\int_0^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}}\tag{1}$$ be the complete elliptic integral of first kind. When $k$ is available from context we usually write $K, K'$ for $K(k), K(k') $ respectively and the variable $q=\exp(-\pi K'/K) $ is called the nome corresponding to $k$.
A lot of functions of the variable $q$ occurring naturally in elliptic function theory can be expressed in terms of the elliptic moduli and integrals. One such relevant function is Dedekind's eta function given by $$\eta(q) =q^{1/24}\prod_{n=1}^{\infty}(1-q^n)\tag{2}$$ Another function needed for our purpose here is the logarithmic derivative of eta function $$P(q)=24q\frac{d}{dq}\log\eta(q)=1-24\sum_{n=1}^{\infty}\frac{nq^n}{1-q^n}\tag{3}$$ We have the closed form evaluations
\begin{align}
\eta(q)&=2^{-1/6}\sqrt{\frac{2K}{\pi}}k^{1/12}k'^{1/3}\tag{4a}\\
\eta(q^2)&=\eta(\exp(-2\pi K'/K)) =2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{4b}
\end{align}
Differentiating the above formulas logarithmically with respect to $q$ and making use of $(3)$ as well as the formula $$\frac{dq} {dk} =\frac{\pi^2q}{2kk'^2K^2}\tag{5}$$ we get $$2P(q^2)-P(q)=\left(\frac{2K}{\pi}\right)^2(1+k^2)$$ Replacing $k$ with $(1-k')/(1+k')$ changes $q$ into $q^2$ and $K$ into $(1+k')K/2$ (Landen transformation) and we get $$2P(q^4)-P(q^2)=\left(\frac{2K}{\pi}\right)^2\frac{1+k'^2}{2}$$ Putting $k=k'=2^{-1/2}$ so that $q=e^{-\pi} $ and elliptic integral $K=K(2^{-1/2})=\Gamma^2(1/4)/(4\sqrt{\pi})$ we get $$P(e^{-4\pi})=\frac{3}{2\pi}+\frac{3\Gamma^4(1/4)}{32\pi^3}\tag{6}$$ (using well known value $P(e^{-2\pi})=3/\pi$).
Next we need a series for $(2K/\pi)^2$ namely $$\left(\frac{2K}{\pi}\right)^{2} = \{1-(kk') ^2\} ^{-1/2}\,_{3}F_{2}\left(\frac{1}{6},\frac{5}{6}, \frac{1}{2};1, 1; \frac{27g^{24}}{(4g^{24} + 1)^{3}}\right)\tag{7}$$ where $g=(2k/k'^2)^{-1/12}$. Using $k=3-2\sqrt{2}$ (singular modulus corresponding to nome $q=e^{-2\pi}$) we get the series mentioned in question $$\sum_{n=0}^{\infty}\frac{(6n)!}{n!^3(3n)!}\left(\frac{1}{287496} \right)^n=\frac{\sqrt{33}\Gamma(1/4)^4}{32\pi^3}$$ The calculation is greatly simplified if we observe that $$g^{24}=8,1-(kk')^2=1-k^2+k^4=k^2(4g^{24}+1)$$ and noting that $$K(k)=\frac{1+2^{-1/2}}{2}K(2^{-1/2})$$ via Landen transformation.
We can write using $(4b)$ $$\eta^4(q^2)=\frac{2^{-4/3}(kk')^{2/3}}{\sqrt{1-k^2+k^4}}\sum_{n=0}^{\infty}\frac{(1/6)_n(5/6)_n(1/2)_n}{(n!)^3}\left(\frac{27g^{24}}{(4g^{24}+1)^3}\right)^n$$ Next task is to differentiate the above series logarithmically with respect to $k$ and this is bit of a computational challenge. In the process we also make use of formula $(5)$. After a lot of tedious algebraic manipulation (readers should attempt verification only if they have a lot of spare time) one can obtain $$P(q^2)=\frac{(1-2k^2)(2-k^2)(1+k^2)}{2(1-k^2+k^4)^{3/2}} \sum_{n=0}^{\infty}\frac{(1/6)_n(5/6)_n(1/2)_n}{(n!)^3}(1+6n)\left(\frac{27g^{24}}{(4g^{24}+1)^3}\right)^n$$ Putting $k=3-2\sqrt{2}$ we get $$P(e^{-4\pi})=\frac{63}{11\sqrt {33}}\sum_{n=0}^{\infty}\frac{(1/6)_n(5/6)_n(1/2)_n}{(n!)^3}(1+6n)\left(\frac{2}{11}\right)^{3n}$$ We had already computed the value of $P(e^{-4\pi})$ as $$P(e^{-4\pi})=\frac{3}{2\pi}+\frac{3}{\sqrt{33}}\sum_{0}^{\infty}\frac{(1/6)_n(5/6)_n(1/2)_n}{(n!)^3}\left(\frac{2}{11}\right)^{3n}$$ and on comparing these values we get $$\frac{11\sqrt{33}}{4\pi}=\sum_{n=0}^{\infty}\frac{(1/6)_n(5/6)_n(1/2)_n}{(n!)^3}(63n+5)\left(\frac{2}{11}\right)^{3n}$$ A major (and more difficult) part of Ramanujan's technique deals with evaluation of $P(q^2) $ in symbolic form as $$P(q^2)=\frac{3}{\pi\sqrt{n} } +\left(\frac{2K}{\pi}\right)^2f_n(k)$$ for $q=e^{-\pi\sqrt{n}}$ where $f_n(k) $ is some complicated algebraic function of $k$ dependent on $n$. Ramanujan gave explicit formulas for many values of $n$ including the case $n=4$ relevant here. We have instead taken a simpler route to evaluate $P(e^{-4\pi})$ using the well known value of $P(e^{-2\pi})$.
I also checked the paper linked in question and I find the claims made by its author regarding Ramanujan's technique being non-rigorous as simply a misconception. The development by Ramanujan in the theory of elliptic and theta functions is fully rigorous and is nothing more than processes of calculus applied in unconventional contexts and involves significant amount of computational labor. But it is a lot easier to understand his ideas and theories compared to the modern approach given by modular forms.
In particular let us observe that Ramanujan did not give details of computation, but the same goes for Borwein brothers and Chudnovsky brothers who didn't give details of their software based calculations. Berndt and his collaborators have been more forthcoming in their papers and give reasonable details of the calculations involved so that others can independently verify it if they have access to the needed software. Ramanujan's calculations were probably done on slate and never stored anywhere but their lack does not imply a lack of rigor.
$\DeclareMathOperator{\cl}{cl}$
$\DeclareMathOperator{\sl}{sl}$
Let $C/\mathbb{Q}$ be the plane curve
$$c^2 + s^2 + c^2 s^2 = 1$$
with distinguished point $O:(c_O,s_O)=(1,0)$. Then $C$ is an elliptic curve, birational by some $\phi: C\rightsquigarrow E$ to the curve $E/\mathbb{Q}$ with Weierstrass form
$$y^2 = x^3 + 4x\text{,}$$
and therefore admits the usual group law $(+,-,O)$. Let $K$ be an algebraic number field, let $\mathfrak{p}$ be a real place of $K$, and let $P:(\gamma,\sigma)$ be a point of $C/K$. Then the question has two parts:
- Can we decide whether $P$ an $N$-torsion point of $C/K$ for some $N$?
- Suppose $P$ is an $N$-torsion point. Let $\Omega$ be the period lattice and $D$ the fundamental parallelogram of the lemniscate elliptic functions. Can we find the unique $t\in\tfrac{1}{N}\Omega \cap D$ such that $$(\gamma_{\mathfrak{p}},\sigma_{\mathfrak{p}})= (\cl t, \sl t)\text{?}$$
And the answer to both parts is "yes". Sketch of an algorithm:
- Compute the torsion subgroup $E(K)_{\text{tors}}$ of $E/K$. This the "hi-tech" part, but it has been possible for a while, basically because we have computable bounds for the size of the torsion subgroup in terms of the properties of $K$, and is implemented in Magma.
- Test membership of $\phi(P)\in E(K)_{\text{tors}}$. If false, then $P$ is not an $N$-torsion point for any $N$, and we are done.
- Otherwise, we know $P$ is an $N$-torsion point for some $N$. But we have a real place $\mathfrak{p}$, so we know that $-1 < \sigma_{\mathfrak{p}} < 1$. And $\sl$ and $\cl$, as elliptic functions, take every value away from ramification points exactly twice. So writing $2\varpi$ for the real period, we know that $$(\gamma_{\mathfrak{p}},\sigma_{\mathfrak{p}})=(\cl \tfrac{2\varpi M}{N},\sl \tfrac{2\varpi M}{N})$$ for a unique value of $0<M<N$. But the left side, and each of the $N-2$ possible values for the right side, are effectively computable; consequently, our estimate for the left side is inconsistent with all but one of the possible values of the right side after computing to sufficient precision. Therefore we can compute the value of $M$, and so have solved the second part of the question.
Examples
$\mathbb{Z}/(8)$-torsion from $\mathbb{Q}(\sqrt{\sqrt{2}-1})$
Let $K=\mathbb{Q}(\alpha)$, where $\alpha=\sqrt{\sqrt{2}-1}$. Then, even if we didn't know already, we could establish that $(\sqrt{\sqrt{2}-1},\sqrt{\sqrt{2}-1})=(\cl \tfrac{M\varpi}{4},\sl\tfrac{M\varpi}{4})$, i.e., $\int_0^{\alpha}\frac{\mathrm{d}s}{\sqrt{1-s^4}}=\tfrac{M\varpi}{4}$, for some integer $M$. That's because we can compute the torsion subgroup to be isomorphic to $\mathbb{Z}/(8)$. Using the Magma calculator
R<z> := PolynomialRing(Integers());
K<a> := NumberField(z^4 + 2*z^2 - 1);
P<c, s, l> := ProjectiveSpace(K, 2);
C := Curve(P, c^2*s^2 + c^2*l^2 + s^2*l^2 - l^4);
O := C ! [0, 1, 1];
E, phi := EllipticCurve(C, O);
EKtors<e>, map := TorsionSubgroup(E);
EKtors;
RationalPoints(Pullback(phi, map(EKtors ! 5)));
yields
Abelian Group isomorphic to Z/8
Defined on 1 generator
Relations:
8*e = 0
{@ (0 : 1 : 1), (a : a : 1), (0 : 1 : 0), (1 : 0 : 0) @}
$\mathbb{Z}/(12)$-torsion from $\mathbb{Q}(\sqrt[4]{\sqrt{12}-3})$
Let $K=\mathbb{Q}(\alpha)$, where $\alpha=\sqrt[4]{\sqrt{12}-3}$. Once again, even if we didn't know already, we could establish that $(\sqrt[4]{\sqrt{12}-3},\tfrac{1}{2}(1+\sqrt{3}-\sqrt[4]{12}))=(\cl \tfrac{M\varpi}{6},\sl\tfrac{M\varpi}{6})$, i.e., $\int_0^{\tfrac{1}{2}(1+\sqrt{3}-\sqrt[4]{12})}\frac{\mathrm{d}s}{\sqrt{1-s^4}}=\tfrac{M\varpi}{6}$, for some integer $M$. That's because we can compute the torsion subgroup to be isomorphic to $\mathbb{Z}/(12)$. Using the Magma calculator again
R<z> := PolynomialRing(Integers());
K<a> := NumberField(z^8 + 6*z^4 - 3);
P<c, s, l> := ProjectiveSpace(K, 2);
C := Curve(P, c^2*s^2 + c^2*l^2 + s^2*l^2 - l^4);
O := C ! [0, 1, 1];
E, phi := EllipticCurve(C, O);
EKtors<e>, map := TorsionSubgroup(E);
EKtors;
RationalPoints(Pullback(phi, map(EKtors ! 10)));
yields
Abelian Group isomorphic to Z/12
Defined on 1 generator
Relations:
12*e = 0
{@ (0 : 1 : 1), (a : 1/4*(-a^6 + a^4 - 5*a^2 + 5) : 1), (0 : 1 : 0), (1 : 0 : 0)
@}
Best Answer
Let us use the more standard notation with $k\in(0,1)$ being elliptic modulus and $k'=\sqrt{1-k^2}$ being the complementary modulus. The elliptic integrals are defined by $$K(k) =\int_0^{\pi/2}\frac {dx} {\sqrt{1-k^2\sin^2x}},E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2x}\,dx\tag{1}$$ The integrals $K(k), K(k'), E(k), E(k') $ are often denoted by $K, K', E, E'$ when the values $k, k'$ are available from context.
Let us now bring other players into this game and we write $q=\exp(-\pi K'/K) $ and call it the nome corresponding to modulus $k$. Ramanujan now makes use of the Dedekind eta function $$\eta(q) =q^{1/24}\prod_{n\geq 1}(1-q^n)\tag{2}$$ and its logarithmic derivative $$P(q) = 24q\frac{d}{dq}\log\eta(q)=1-24\sum_{n\geq 1}\frac{nq^n}{1-q^n}\tag{3}$$ We have the identity $$\eta(q^2)=2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{4}$$ which will be used in combination with the hypergeometric identity $$ \left(\frac{2K}{\pi}\right)^{2} = (1 - 2k^{2})^{-1}\,_{3}F_{2}\left(\frac{1}{4},\frac{3}{4},\frac{1}{2}; 1, 1; -\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2}\right)\tag{5}$$ where $G=G(k) =(2kk') ^{-1/12}$ is one of Ramanujan's class invariant. Using these identities we get $$\eta^4(q^2)=\frac{2^{-4/3}(kk')^{2/3}}{1-2k^2} \,_{3}F_{2}\left(\frac{1}{4},\frac{3}{4},\frac{1}{2}; 1, 1; -\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2}\right)\tag{6}$$ Next formidable step is to logarithmically differentiate the above identity with respect to $k$ to get a formula for $P(q^2)$. The calculations can be verified with hand and some patience and we write a few steps here. First we have $$P(q^2)=3q\frac{d}{dq}\log\eta^4(q^2)=\frac{3kk'^2}{2}\left(\frac{2K}{\pi}\right)^2\frac{d}{dk}\log\eta^4(q^2)$$ which further leads us to $$P(q^2)=\frac{3kk'^2}{2}\left(\frac{2K}{\pi}\right)^2\frac{d}{dk}\left(\log (2^{-4/3}(kk')^{2/3})+\log\left(\frac {2K}{\pi}\right) ^2\right) $$ Then we get $$P(q^2)=\left(\frac{2K}{\pi}\right) ^2(1-2k^2)+\frac{3kk'^2}{2}\frac{d}{dk}\left(\frac{2K}{\pi}\right)^2$$ Using $(5)$ we now get $$P(q^2)=\left(\frac{2K}{\pi}\right) ^2\cdot\frac{1+2k^2k'^2}{1-2k^2}+\frac{3kk'^2}{2(1-2k^2)}\frac{d}{dk}\,_{3}F_{2}\left(\frac{1}{4},\frac{3}{4},\frac{1}{2}; 1, 1; -\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2}\right)$$ Switching to sigma notation we get $$P(q^2)=\frac{1+2k^2k'^2}{(1-2k^2)^2}\sum_{n\geq 0}(-1)^n\cdot\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2n}+\frac{3kk'^2}{2(1-2k^2)}\cdot\frac{f'(k)}{f(k)}\sum_{n\geq 0}(-1)^n\cdot\frac{n(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2n}$$ where $$f(k)= \left(\frac{G^{12} - G^{-12}}{2}\right)^{-2}$$ We have $$\frac{f'(k)} {f(k)} =(\log f(k)) '=\frac{2(1-2k^2)}{kk'^2}\cdot\frac{G^{12}+G^{-12}}{G^{12}-G^{-12}}$$ Thus we finally obtain $$P(q^2)=\frac{1+2k^2k'^2}{(1-2k^2)^2}\sum_{n\geq 0}(-1)^n\cdot\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2n}+\frac{3(G^{12}+G^{-12})}{G^{12}-G^{-12}}\sum_{n\geq 0}(-1)^n\cdot\frac{n(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\left(\frac{G^{12} - G^{-12}}{2}\right)^{-2n}\tag{7} $$ Next we put $q=e^{-5\pi}$ in above relation and corresponding values $$2k^2=1-\sqrt{1-\phi^{-24}},2k'^2=1+\sqrt{1-\phi^{-24}}, G=(4k^2k'^2)^{-1/24}=\phi$$ (where $\phi=(1+\sqrt{5})/2$ is golden ratio) and observe that $$G^{12}-G^{-12}=F_{12}\phi+F_{11}-(-F_{12}\phi+F_{13})=2F_{12}\phi-F_{12}=F_{12}\sqrt{5}=144\sqrt{5}$$ (where $F_n$ is the Fibonacci sequence) and $$G^{12}+G^{-12}=F_{11}+F_{13}=322$$ and $$\frac{1+2k^2k'^2}{(1-2k^2)^2}=\frac{2F_{24}\phi+2F_{23}+1}{2F_{24}\phi+2F_{23}-2}=\frac{30912\phi+19105}{30912\phi+19104}$$ The above ratio can be simplified (using technique in this answer) to $$\frac{120+161\sqrt{5}}{480}$$ Thus the equation $(7)$ can be rewritten as $$P(q^2)=\frac{120+161\sqrt{5}}{480}\sum_{n\geq 0}(-1)^n\cdot\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\frac{1}{25920^n}+\frac{644\sqrt{5}}{480}\sum_{n\geq 0}(-1)^n\cdot\frac{n(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\frac{1}{25920^n}\tag{8}$$ From equation $(15)$ of this answer dealing with modular equations and related identities we have $$P(q^{2})=\frac{3}{5\pi}+\left(\frac{2K}{\pi}\right) ^26(5-3\phi)\sqrt{3\phi-4}$$ or $$ P(q^{2})=\frac{3}{5\pi}+\frac{6(5-3\phi)\sqrt{3\phi-4}} {\sqrt{1-\phi^{-24}}} \sum_{n\geq 0} (-1)^n\cdot\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\frac{1}{25920^n}$$ and this simplifies to $$P(q^{2})=\frac{3}{5\pi}+\frac{120+120\sqrt{5}} {480} \sum_{n\geq 0} (-1)^n\cdot\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\frac{1}{25920^n}\tag{9}$$ Comparing equations $(8),(9)$ we get $$\frac{288\sqrt{5}}{5\pi}=\sum_{n\geq 0}(-1)^n\cdot\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}\cdot\frac{41+644n}{25920^n} $$