Ramanujan's fastest series for 1/π(PI)

In his paper Modular equations and approximations to $\pi$ Ramanujan gave the following spectacular series for $1/\pi$: $$\frac{1}{2\pi\sqrt{2}} = \frac{1103}{99^{2}} + \frac{27493}{99^{6}}\frac{1}{2}\frac{1\cdot 3}{4^{2}} + \frac{53883}{99^{10}}\frac{1\cdot 3}{2\cdot 4}\frac{1\cdot 3\cdot 5\cdot 7}{4^{2}\cdot 8^{2}} + \cdots\tag{1}$$ without any proof. The series converges very fast with each term giving almost 8 digits of the final sum. It was proved by Borwein brothers in 1987 with some calculations assisted via computer software. And most recently in 2025 it was proved by Jesus Guillera using MAPLE software. There is no known proof using hand calculation available so far. Using ideas from a paper by Pablo F. Refolio, an independent researcher from Spain, I will present a proof which involves calculations via hand, pen and paper.

Generalized View of Modular Equations

Let us repeat for the benefit of readers a few ideas from the theory of elliptic integrals and modular equations and theta functions.

Let $k\in(0,1)$ be the elliptic modulus, $k'=\sqrt{1-k^2}$ be the complementary modulus and then we define complete elliptic integrals of first and second kinds as $$K(k) =\int_0^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}} , \, E(k) =\int_{0}^{\pi/2}\sqrt {1-k^2\sin^2x}\,dx\tag{2}$$ respectively. The integrals $K(k), K(k'), E(k), E(k') $ are usually denoted by $K, K', E, E'$ when the values of $k, k'$ are available from context and these satisfy the Legendre's identity $$KE'+K'E-KK'=\frac{\pi} {2}\tag{3}$$ Often a related modulus $l$ and its complementary modulus $l'=\sqrt{1-l^2}$ is used and we denote $K(l), K(l') $ by $L, L'$ respectively.

It can be proved using definition of elliptic integrals that the ratio $K'/K$ decreases strictly as $k$ increases and maps $(0,1)$ to the set of positive real numbers (or in short $(0,\infty)$). Therefore if modulus $k\in(0,1)$ and a positive real number $n$ is given the ratio $nK'/K$ is positive as well and a unique positive modulus $l\in(0,1)$ exists such that $$\frac{L'} {L} =n\frac{K'} {K}\tag{4} $$ Jacobi proved in his Fundamenta Nova that if $n$ is a positive rational number then the relationship between $k, l$ induced by equation $(4)$ is algebraic, that is, there exists a polynomial $P(x, y) $ with integer coefficients such that $P(k, l) =0$. The equation $P(k, l) =0$ is called a modular equation of degree $n$. These equations are usually not given using polynomials (because they can involve large number of terms and become clumsy to handle) but using some algebraic functions involving $k, k', l, l'$.

It is almost a piece of magic that given the values of integrals $K, K'$ one can get the values of moduli $k, k'$ using functions of another associated variable $q$ called nome (corresponding to modulus $k$) given by $$q=\exp\left(-\pi\frac{K'} {K} \right) \tag{5}$$ and often written shortly as $q=e^{-\pi K'/K} $. To get $k, k'$ in terms of $q$ we need theta functions of Jacobi in simplified form as \begin{align} \vartheta_2(q)&=\sum_{n\in\mathbb{Z}}q^{(n+(1/2))^2}=2q^{1/4}\psi(q^2)\tag{6a} \\ \vartheta_3(q)&=\sum_{n\in\mathbb{Z}}q^{n^2}\tag{6b}\\ \vartheta_4(q)&=\vartheta_3(-q)\tag{6c} \end{align} where $$\psi(q) =\sum_{n=0}^{\infty} q^{n(n+1)/2}\tag{7}$$ is a theta function defined by Ramanujan. The series for these theta functions converge for all complex values of $q$ with $|q|<1$ and hence they are defined within the unit disk with center at origin in the complex plane. Using these theta functions one can prove with significant effort that $$k=\frac{\vartheta_2^2(q)} {\vartheta_3^2(q)},\,k'=\frac{\vartheta_4^2(q)}{\vartheta_3^2(q)}, \,K=\frac{\pi} {2}\vartheta_3^2(q)\tag{8}$$ If we observe equations $(4), (5)$ closely we note that for a modular equation of degree $n$ if $q$ is the nome corresponding to modulus $k$ then $q^n$ is the nome corresponding to modulus $l$ and a modular equation of degree $n$ is essentially an algebraic relationship between theta functions of $q$ and $q^n$ which holds for all values of $q$ inside the unit disk centered at origin due to analytic nature of these theta functions.

To deal with fractional powers of $q$ involved in definition of theta function $\vartheta_2$ it is best to define theta functions in terms of another variable $\tau$ with $$q=\exp(\pi i\tau) \tag{9}$$ The restriction of $q$ with $|q|<1$ forces us to consider the values of $\tau$ in upper half of the complex plane above real axis or in other words we consider $\tau$ to have positive imaginary part.

Thus starting with real values of moduli $k, l $ in $(0,1)$ and a given positive rational number $n$ one can obtain a modular equation as an algebraic relationship between $k, l$ and thereby arrive at an algebraic relationship between theta functions of $q$ and $q^n$ which holds for all $q$ inside unit disk centered at origin. Equivalently we have a relation between theta functions of $\tau$ and $n\tau$ which holds for all $\tau$ with positive imaginary part. It is this generalized view of modular equations we need here in our proof for Ramanujan's fastest series for $1/\pi$. It should also be observed that if $\tau$ is purely imaginary then $k, k', q$ are real and lie in $(0,1)$ and that is the usual setting we started with. But if $\tau$ has non-zero real part the values of $k, k', q$ are no longer restricted to interval $(0,1)$ with only restriction being $|q|<1$.

Since theta functions can be viewed as both functions of nome $q$ and parameter $\tau$ we will use the functional notation by writing $\vartheta_j(q) $ and $\vartheta_j(\tau) $ interchangeably. In the light of equation $(8)$ the moduli $k, k'$ and the integral $K$ can also be viewed as functions of $q$ as well functions of $\tau$ and we will use the functional notation like $$k(q), k'(q), K(q), k(\tau), k'(\tau), K(\tau), K(k)$$ as and when required. We will use the explicit notation $d/dk, d/dq$ for derivatives and avoid the prime notation so that $k'(q) $ is not the derivative of $k(q) $ but rather the moduli $k'$ expressed as a function of $q$.

Let us also note the fundamental transformation formulas for theta functions \begin{align} \vartheta_2(-1/\tau)&=(-i\tau)^{1/2}\vartheta_4(\tau) \tag{10a}\\ \vartheta_3(-1/\tau)&=(-i\tau)^{1/2}\vartheta_3(\tau) \tag{10b}\\ \vartheta_4(-1/\tau)&=(-i\tau)^{1/2}\vartheta_2(\tau) \tag{10c}\\ \end{align} which lead us to $$k'(-1/\tau)=k(\tau)$$ We end our description of elliptic integrals and theta functions with mention of Landen transformation which says that $$K(2\tau)=\frac{1+k'(\tau)}{2}K(\tau), k(2\tau)=\frac{1-k'(\tau)}{1+k'(\tau)}\tag{11a}$$ These formulas above can be inverted as well to get $$K(\tau/2)=(1+k(\tau))K(\tau),\,k(\tau/2)=\frac{2\sqrt{k(\tau)}}{1+k(\tau)}\tag{11b}$$ where $\sqrt{k(\tau)} =\vartheta_2(\tau)/\vartheta_3 (\tau) $.

Ramanujan's function $P(q)$

We need another ingredient here called Dedekind's eta function defined by $$\eta(q) =q^{1/24}\prod_{j=1}^{\infty}(1-q^j)\tag{12}$$ Using product representation of theta functions one can prove that $$\eta(q^2)=2^{-1/3}\sqrt {\frac{2K}{\pi}} (kk') ^{1/6}\tag{13}$$ Ramanujan defined a function $P(q) $ as $$P(q) =24q\frac{d}{dq}\log\eta(q)=1-24\sum_{j=1}^{\infty} \frac{jq^j} {1-q^j}\tag{14}$$ Using the derivative formula of elliptic integrals $$\frac{dK} {dk}=\frac{E-k'^2K}{kk'^2}$$ one can prove that $$\frac{dq} {dk} =\frac{\pi^2q}{2kk'^2K^2}\tag{15}$$ and then using definition of $P(q) $ one can prove that $$P(q^2)=\left(\frac{2K}{\pi}\right)^2\left(\frac{3E}{K}+k^2-2\right)\tag{16}$$ With some more effort (Landen transformation) one can also show that $$P(q) =\left(\frac{2K}{\pi}\right)^2\left(\frac{6E}{K}+k^2-5\right)\tag{17} $$ and thus $$2P(q^2)-P(q)=\left(\frac{2K}{\pi}\right)^2(1+k^2)\tag{18}$$ Using definition of $P(q) $ it is also possible to prove that $$P(q^2)=\left(\frac{2K}{\pi}\right)^2(1-2k^2)+\frac{3kk'^2}{2}\frac{d} {dk} \left(\frac{2K}{\pi}\right)^2\tag{19}$$ In terms of parameter $\tau$ we define corresponding to $P(q) $ an Eisenstein series $$E_2(\tau) =1-24\sum_{j=1}^{\infty}\frac{je^{2\pi ij\tau}} {1-e^{2\pi ij\tau}}=P(q^2) \tag{20}$$ which satisfies the obvious property $E_2(\tau+1)=E_2(\tau)$ and the highly non-obvious identity $$E_2(-1/\tau)=\tau^2E_2(\tau)-\frac{6i\tau}{\pi}\tag{21}$$ The $\pi$ in denominator on right hand side of the above identity is key to establishing the series for $1/\pi$.

Ramanujan used his deep skills of algebraic manipulation to find expressions for $$nP(q^{2n})-P(q^2)=nE_2(n\tau)-E_2(\tau)$$ for various positive integers $n$ using modular equations of degree $n$. If $k, l$ are moduli corresponding to nomes $q, q^n$ respectively then he presented his results in the form $$nP(q^{2n})-P(q^2)=\frac{4KL}{\pi^2} F_n(k, l) $$ where $F_n(k, l) $ is a complicated algebraic function. These expressions are essentially representations for $nE_2(n\tau)-E_2(\tau)$ in terms of theta functions of $\tau$ and $n\tau$ and are valid for all $\tau$ with positive imaginary part.

In his paper Ramanujan used these expressions to get various series for $1/\pi$ but in the examples he provided he used purely imaginary values of $\tau$ or in other words used moduli and nomes in interval $(0,1)$. Refolio's paper uses complex values of $\tau$ in a very smart way to reduce the degree of modular equations involved and thereby simplifies calculations considerably so that they can be performed using hand, pen and paper. We illustrate his idea and its application to Ramanujan's fastest series for $1/\pi$ next.

Using Parameter $\tau$ with non-zero Real Part

Let $p$ be an odd positive integer and let $\tau=i\sqrt{p} +1$ so that $$(\tau-1)^2=-p$$ or $$\tau^2-2\tau=-(p+1)$$ We write $p+1=2n$ where $n$ is a positive integer and then $$\tau-2=n(-2/\tau)$$ and hence moduli corresponding to $\tau-2$ and $-2/\tau$ satisfy a modular equation of degree $n$. In order to use the modular equation effectively we need to get expressions for all the moduli and integrals involved in such modular equations. Let us then note that $$k(\tau-2)=k(i\sqrt{p}-1)$$ and using formula $$k=\frac{4q^{1/2}\psi^2(q^2)} {\vartheta_3^2(q)} $$ we note that if $q$ is replaced by $qe^{-i\pi}=-q$ the expression for $k$ changes to $$ \frac{-4iq^{1/2}\psi^2(q^2)}{\vartheta_3^2(-q)}=-i\cdot\frac{4q^{1/2}\psi^2(q^2)/\vartheta_3^2(q) } {\vartheta_4^2(q)/\vartheta_3^2(q)}$$ and thus we get $$k(\tau-2)=k(i\sqrt{p} - 1)=-i\frac{k_p}{k'_p}$$ where $k_p$ is singular moduli correspondingto nome $q=e^{-\pi\sqrt{p}} $. The expression above for $k(\tau-2)$ can be written in terms of Ramanujan's class invariant $G_p=(2k_pk'_p) ^{-1/12}$ as $$k(\tau-2)=i(\sqrt{G_p^{24}-1}-G_p^{12})$$ which is evaluated easily if $G_p$ is known. Let us also note that $$k'(\tau-2)=1/k'_p, K(\tau-2)=k'_pK(k_p)$$ Next we need to evaluate moduli and integrals for parameter $(-2/\tau)$. This we do by first dealing with parameter $-1/\tau$ and then applying Landen transformation. We have $$k'(-1/\tau)=k(\tau)=k(i\sqrt{p}+1)=i\frac{k_p} {k'_p} $$ and then $$k(-2/\tau)=\frac{1-k'(-1/\tau)}{1+k'(-1/\tau)}=\frac{1-ik_p/k'_p}{1+ik_p/k'_p} =(k'_p-ik_p)^2$$ and the above can be written as $$\sqrt{1-G_p^{-24}}-iG_p^{-12}$$ We also have $$K(-1/\tau)=(-i\tau)K(\tau)=(\sqrt {p} - i) K(i\sqrt {p} +1)=(\sqrt {p} - 1)k'_pK(k_p)$$ and then using Landen transformation we have $$K(-2/\tau)=\frac{1+k'(-1/\tau)}{2}K(-1/\tau)$$ which simplifies to $$\frac{\sqrt {p} - i} {2}(k'_p+ik_p)K(k_p)$$ Refolio uses the above approach for $p=37$ with $n=(p+1)/2=19$ in combination with Ramanujan's formula for $19P(q^{38})-P(q^2)$ to prove the following marvelous series (given in Ramanujan's paper) for $1/\pi$ $$\frac{4}{\pi}=\frac{1123}{882}-\frac{22583}{882^{3}}\cdot\frac{1}{2}\cdot\frac{1\cdot 3}{4^{2}}+\frac{44043}{882^{5}}\cdot\frac{1\cdot 3}{2\cdot 4}\cdot\frac{1\cdot 3\cdot 5\cdot 7}{4^{2}\cdot 8^{2}}-\dots$$ Proof in Refolio's paper is the first proof for above series which involves calculations done by hand. All other known proofs use computer software for algebraic manipulation.

We adapt the above approach to deal with even positive values of $p$ by taking parameter $\tau=i\sqrt {p} +2$ so that $$\tau^2-4\tau=-(p+4)$$ We let $p+4=2n$ and then $$\tau-4=n(-2/\tau)$$ so that there is a modular equation of degree $n$ connecting moduli corresponding to parameters $\tau-4=i\sqrt{p} - 2$ and $-2/\tau$.

We illustrate the approach by taking $p=58$ so that $n=(p+4)/2=31$ and then $$i\sqrt {58} - 2=31\left(\frac{-2} {i\sqrt{58} +2}\right) $$ We first evaluate moduli and integrals corresponding to $\tau-4$ and do the same for $-2/\tau$. We have $$k(\tau-4)=k(i\sqrt{p}-2)$$ and in the expression $k=4q^{1/2}\psi^2(q^2)/\vartheta_3 ^2(q)$ if we replace $q$ with $qe^{-2i\pi}$ and $q^{1/2}$ with $q^{1/2}e^{-i\pi} =-q^{1/2}$ then we get $k=-k_p$ and further $k'=k'_p$. We have $$K(\tau-4) =K(i\sqrt {p} - 2)=K(k_p)$$ To get moduli and integrals for parameter $-2/\tau$ we note that $$k'(-1/\tau)=k(\tau)=k(i\sqrt{p}+2)=-k_p$$ and then $$k(-2/\tau)=\frac{1-k'(-1/\tau)}{1+k'(-1/\tau)}=\frac{1+k_p}{1-k_p}$$ Further $$K(-1/\tau)=(-i\tau) K(\tau) =(\sqrt{p} - 2i)K(i\sqrt {p} +2)=(\sqrt {p} - 2i)K(k_p)$$ and thus $$K(-2/\tau)=\frac{1+k'(-1/\tau)}{2}K(-1/\tau)=\frac{1+k(\tau)}{2}K(-1/\tau)$$ which simplifies to $$\frac{\sqrt{p} - 2i}{2}(1-k_p)K(k_p)$$ We next reset notation and let $k, l$ denote moduli corresponding to parameters $$\tau=-\frac{2}{i\sqrt{58}+2},31\tau=i\sqrt {58}-2$$ respectively and let $K, L$ be corresponding elliptic integrals for these moduli. Then we have $$k=\frac{1+k_{58}}{1-k_{58}},l=-k_{58},k'=\frac{2i\sqrt {k_{58}}}{1-k_{58}}, l'=k'_{58},\\ K=\frac{\sqrt{58}-2i}{2}(1-k_{58})K(k_{58}),L=K(k_{58})\tag{22}$$ With some more effort we derive $$ \sqrt{l}=-i\sqrt{k_{58}}, l^{1/4}= \frac{1-i}{\sqrt{2}}(k_{58})^{1/4}, (k')^{1/2}=\frac{(1+i)\sqrt[4]{k_{58}}}{\sqrt{1-k_{58}}},\\(k')^{1/4}=\left(\sqrt{\frac{\sqrt{2}+1}{2}}+i\sqrt{\frac{\sqrt{2}-1}{2}}\right)\frac{\sqrt[8]{k_{58}}}{\sqrt[4]{1-k_{58}}}\tag{23}$$ The expressions for $k^{1/2}, k^{1/4}, (l') ^{1/2},(l')^{1/4}$ are just square and fourth roots of real positive numbers $k,l'$ given above. We then have modulus $l$ of degree $31$ over $k$.

The above set of values will be used in Ramanujan's formula $$31E_2(31\tau)-E_2(\tau)=31P(q^{62})-P(q^2)= \frac{4K(k)K(l)}{\pi^2}F_{31}(l,k)$$ with $$F_{31}(l,k) = 9(1+kl+k'l') + 12(\sqrt{kl}+\sqrt{k'l'}+\sqrt{kk'll'})\\ \, \, \, \, \, \, \, \, \, -12(kk'll')^{1/4}(1+(kl)^{1/4}+(k'l')^{1/4})\tag{24}$$ Ramanujan's formula above will be used to evaluate $P(q^{62})=P(e^{-2\pi\sqrt{58}})$ and to evaluate $P(q^2) = E_2(\tau)$ we use equation $(21)$ to get $$E_2(\tau)=(1/\tau)^2E_2(-1/\tau)+\frac{6i}{\pi\tau}$$ and observe that $E_2(-1/\tau)=P(e^{-\pi\sqrt{58}})$ so that $$P(q^2)=E_2(\tau)=-\frac{27-2i\sqrt{58}}{2}P(e^{-\pi\sqrt{58}})-\frac{3i(i\sqrt{58}+2)}{\pi}$$ and thus we have $$31P(e^{-2\pi\sqrt{58}})+\frac{27-2i\sqrt{58}}{2}P(e^{-\pi\sqrt{58}})+\frac{3i(i\sqrt{58}+2)}{\pi}=\frac{4K(k)K(l)}{\pi^2}F_{31}(l,k)$$ Next we use the equation $(18)$ to get $$2P(e^{-2\pi\sqrt{58}})-P(e^{-\pi\sqrt{58}})=\left(\frac{2K(k_{58})}{\pi}\right)^2(1+k_{58}^2)$$ Using last two equations and values of $K(k),K(l)$ we get $$31P(e^{-2\pi\sqrt{58}})+\frac{27-2i\sqrt{58}}{2}\left(2P(e^{-2\pi\sqrt{58}})-\left(\frac{2K(k_{58})}{\pi}\right)^2(1+k_{58}^2)\right)+\frac{3i(i\sqrt{58}+2)}{\pi}\\=\left(\frac{2K(k_{58})}{\pi}\right)^2\frac{\sqrt{58}-2i}{2}(1-k_{58})F_{31}(l,k)$$ or $$(58-2i\sqrt{58})P(e^{-2\pi\sqrt{58}})+\frac{3i(i\sqrt{58}+2)}{\pi}\\=\left(\frac{2K(k_{58})}{\pi}\right)^2\left(\frac{27-2i\sqrt{58}}{2}(1+k_{58}^2) + \frac{\sqrt{58}-2i}{2}(1-k_{58})F_{31}(l,k)\right)$$ or $$P(e^{-2\pi\sqrt{58}})-\frac{3}{\pi\sqrt{58}}\\=\left(\frac{2K(k_{58})}{\pi}\right)^2\left(\frac{27-2i\sqrt{58}}{2(58-2i\sqrt{58})}(1+k_{58}^2) + \frac{\sqrt{58}-2i}{2(58-2i\sqrt{58})}(1-k_{58})F_{31}(l,k)\right)$$ or $$P(e^{-2\pi\sqrt{58}})-\frac{3}{\pi\sqrt{58}}\\=\left(\frac{2K(k_{58})}{\pi}\right)^2\left(\frac{58-2i\sqrt{58}}{4\cdot 58}(1+k_{58}^2) + \frac{1-k_{58}}{2\sqrt{58}}F_{31}(l,k)\right)\\= \left(\frac{2K(k_{58})}{\pi}\right)^2(A+iB)\text{ (say)}\tag{25}$$ and let $$a+ib = (1-k_{58})F_{31}(l,k)$$ Then we know that $B=0$ and $$A=\frac{1+k_{58}^2}{4}+\frac{a}{2\sqrt{58}}\tag{26}$$ The challenge now is to evaluate the real part $a$ of $(1-k_{58})F_{31}(l,k)$ given the complicated formula for $F_{31}(l,k)$ and the values related to moduli $l,k$.

Calculation of Real Part of $(1-k_{58})F_{31}(l,k)$

I have performed the calculations related to each term of $(1-k_{58})F_{31}(l,k)$ and tried to put my results in terms of $g_{58}^6k_{58}$.

Let us display the calculations in bit detail and we write $$(1-k_{58})F_{31}(l,k)=S+T-(U+V+W) $$ where \begin{align} S&=9(1-k_{58})(1+kl+k'l')\notag\\ T&=12(1-k_{58})(\sqrt{kl}+\sqrt{k'l'}+\sqrt{kk'll'})\notag\\ U&=12(1-k_{58})(kk'll')^{1/4}\notag\\ V&=12(1-k_{58})(kl)^{1/2}(k'l')^{1/4}\notag\\ W&=12(1-k_{58})(kl)^{1/4}(k'l')^{1/2}\notag \end{align} We have used the value $$g_{58}=\sqrt{\frac{\sqrt{29}+5}{2}}$$ and the following formulas in our calculations \begin{align} k_p&=g_p^6(\sqrt{g_p^{12}+g_{p}^{-12}}-g_p^6)\notag\\ 1/k_p&=g_p^6(\sqrt{g_p^{12}+g_{p}^{-12}}+g_p^6)\notag\\ (k'_p\sqrt{k_p})^n&=(\sqrt{2}\cdot g_p^6k_p)^n\notag\\ k_{58}&=(\sqrt{2}-1)^6(13\sqrt{58}-99)\notag\\ 1+k_{58}&=(\sqrt{2}-1)^6\sqrt{2}(13\sqrt{29}+70)\notag\\ g_{58}^6&= 13\sqrt{29}+70\notag\\ g_{58}^6-g_{58}^{-6}&=140\notag\\ g_{58}^{12}+g_{58}^{-12}&=2\cdot 99^2\notag\\ g_{58}^{12}-g_{58}^{-12}&=4\cdot 13\cdot 70\sqrt{29}\notag \end{align} We start with \begin{align} \Re(S) & =\Re\left\{9(1-k_{58})\left(1+\frac{1+k_{58}}{1-k_{58}}\cdot(-k_{58})+\sqrt{\frac{1+k_{58}}{1-k_{58}}}\cdot(-i\sqrt{k_{58}})\right)\right\}\notag\\ &=9(1-k_{58}- k_{58}(1+k_{58})) \notag\\ &=9(1-k_{58}^2-2k_{58})\notag\\ &=18k_{58}\left(\frac{{k'}_{58}^2}{2k_{58}}-1\right)\notag\\ &=18k_{58}(g_{58}^{12}-1)\notag\\ &=18g_{58}^6k_{58}(g_{58}^6-g_{58}^{-6})\notag\\ &=18\cdot 140g_{58}^6k_{58}\notag \end{align} Next we have \begin{align} \Re(T) &=\Re\left\{12(1-k_{58})\left(\sqrt{\frac{1+k_{58}}{1-k_{58}}}\cdot(-i\sqrt{k_{58}})+(1+i)\frac{\sqrt[4]{k_{58}}}{\sqrt{1-k_{58}}}\cdot\sqrt{k'_{58}}\right. \right. \notag\\ &\left. \left. \, \, \, \,+ \sqrt{\frac{1+k_{58}}{1-k_{58}}}\cdot(-i\sqrt{k_{58}})\cdot(1+i)\frac{\sqrt[4]{k_{58}}}{\sqrt{1-k_{58}}}\cdot\sqrt{k'_{58}}\right)\right\} \notag\\ &=12\cdot 2^{1/4}g_{58}^3\left(\sqrt{k_{58}(1-k_{58})}+k_{58}\sqrt{1+k_{58}}\right)\notag\\ &=12\cdot 2^{1/4}g_{58}^3\left(\sqrt{\frac{{k'}_{58}^2k_{58}} {1+k_{58}} }+k_{58}\sqrt{(\sqrt{2}-1)^6\sqrt{2}g_{58}^6} \right)\notag\\ &=12\cdot 2^{1/4}g_{58}^3\left(\sqrt{\frac{2g_{58}^{12}k_{58}^2} {(\sqrt{2}-1)^6\sqrt{2}g_{58}^6} }+k_{58}\sqrt{(\sqrt{2}-1)^6\sqrt{2}g_{58}^6} \right)\notag\\ &=12\cdot 2^{1/4}g_{58}^3k_{58}\left(\sqrt{\frac{\sqrt{2}g_{58}^6} {(\sqrt{2}-1)^6} }+\sqrt{(\sqrt{2}-1)^6\sqrt{2}g_{58}^6} \right)\notag\\ &=12\cdot 2^{1/2}g_{58}^6k_{58}\left((\sqrt{2}+1)^3+(\sqrt{2}-1)^3\right)\notag\\ &=12\sqrt{2}g_{58}^6k_{58}\cdot 2\sqrt {2}(2(2+1)-1) \notag\\ &= 240g_{58}^6k_{58}\notag \end{align} Next we deal with \begin{align} \Re(U) &=\Re\left[12(1-k_{58})\left\{\sqrt[4]{\frac{1+k_{58}}{1-k_{58}}}\left(\sqrt{\frac{\sqrt{2}+1}{2}}+i\sqrt{\frac{\sqrt{2}-1}{2} }\right)\frac{\sqrt[8]{k_{58}}}{\sqrt[4]{1-k_{58}}}\right. \right. \notag\\ &\left. \left. \, \, \, \, \cdot\frac{1-i}{\sqrt{2}}\sqrt[4]{k_{58}}\sqrt[4]{k'_{58}}\right\}\right]\notag\\ &=6\Re\left[(1-k_{58})\left\{\sqrt[4]{\frac{1+k_{58}}{1-k_{58}}}\left(\sqrt{2}\sqrt{\sqrt{2}+1}-i\sqrt{2}\sqrt{\sqrt{2}-1 }\right)\frac{\sqrt[8]{k_{58}}}{\sqrt[4]{1-k_{58}}}\right. \right. \notag\\ &\left. \left. \, \, \, \, \cdot\sqrt[4]{k_{58}}\sqrt[4]{k'_{58}}\right\}\right]\notag\\ &=6\sqrt{2}\sqrt{\sqrt{2}+1}\left\{\sqrt[4]{(1+k_{58})(1-k_{58})}\sqrt[8]{k_{58}}\sqrt[4]{1-k_{58}}\cdot\sqrt[4]{k_{58}}\sqrt[4]{k'_{58}}\right\}\notag\\ &=6\sqrt{2}\sqrt{\sqrt{2}+1}\sqrt{k'_{58}}\sqrt[8]{k_{58}}\sqrt[4]{1-k_{58}}\cdot\sqrt[4]{k_{58}}\sqrt[4]{k'_{58}}\notag\\ &=6\sqrt{2}\sqrt{\sqrt{2}+1}(k'_{58}\sqrt{k_{58}})^{3/4}\sqrt[4]{1-k_{58}}\notag\\ &=6\sqrt{2}\sqrt{\sqrt{2}+1}(\sqrt{2}g_{58}^6k_{58})^{3/4}\sqrt[4]{1-k_{58}}\notag\\ &=12g_{58}^6k_{58}\left(\frac{(\sqrt{2}+1)^2(1-k_{58})}{\sqrt{2}g_{58}^6k_{58}}\right)^{1/4}\notag\\ &=12g_{58}^6k_{58}\left(\frac{(\sqrt{2}+1)^2{k'} _{58}^2}{\sqrt{2}g_{58}^6k_{58}(1+k_{58})} \right)^{1/4}\notag \\ &=12g_{58}^6k_{58}\left(\frac{(\sqrt{2}+1)^2\cdot 2g_{58}^{12}}{\sqrt{2}g_{58}^6(1+k_{58})}\right)^{1/4}\notag \\ &=12g_{58}^6k_{58}\left(\frac{(\sqrt{2}+1)^2\cdot \sqrt{2}g_{58}^6}{1+k_{58}}\right)^{1/4}\notag \\ &=12g_{58}^6k_{58}\left(\frac{(\sqrt{2}+1)^2\cdot \sqrt{2}g_{58}^6}{(\sqrt{2}-1)^6\sqrt{2}g_{58}^6}\right)^{1/4}\notag \\ &=12(3+2\sqrt{2})g_{58}^6k_{58}\notag \end{align} A similar calculation as above shows that $$\Re(V) = 12(3-2\sqrt {2})g_{58}^6k_{58}$$ so that $$\Re(U+V) =72g_{58}^6k_{58}$$ Finally we have \begin{align} \Re(W)&=12\Re\left\{(1-k_{58})\sqrt[4]{\frac{1+k_{58}}{1-k_{58}}}\cdot\frac{1-i}{\sqrt{2}}\sqrt[4]{k_{58}}\right. \notag\\ &\, \, \, \, \left. \cdot(1+i)\frac{\sqrt[4]{k_{58}}}{\sqrt{1-k_{58}}}\sqrt{k'_{58}}\right\}\notag \\ &=12\sqrt{2}k'_{58}\sqrt{k_{58}}\notag\\ &=24g_{58}^6k_{58}\notag \end{align} We thus get the real part $a$ of $(1-k_{58})F_{31}(l,k)$ as $$a=18\cdot 140\cdot g_{58}^6k_{58}+240\cdot g_{58}^6k_{58}-72\cdot g_{58}^6k_{58}-24\cdot g_{58}^6k_{58}$$ or $$a= 6\cdot 444\cdot g_{58}^6k_{58}$$ We then get $$A=\frac{1+k_{58}^2}{4}+\frac{6\cdot 444\cdot g_{58}^6k_{58}}{2\sqrt{58}}$$ and $$P(e^{-2\pi\sqrt {58}})-\frac{3}{\pi\sqrt{58}}=\left(\frac{2K(k_{58})}{\pi}\right)^2\left(\frac{1+k_{58}^2}{4}+\frac{6\cdot 444\cdot g_{58}^6k_{58}}{2\sqrt{58}}\right)\tag{27}$$ It can be proved using definition of Borwein's alpha function that $$\alpha(p)=\sqrt{p}\cdot\frac{1+k_p^2-A_p}{3}$$ where $$P(e^{-2\pi\sqrt{p}})-\frac{3}{\pi\sqrt{p}}=\left(\frac{2K(k_p)}{\pi}\right)^2A_p$$ In our case we then have $A_{58}=A$ and $$\alpha(58)=\sqrt{58}\cdot\frac{1+k_{58}^2}{4}-\frac{a}{6}$$ Using $$1+ k_{58}^2=2-{k'}_{58}^2= 2k_{58}\left(\frac{1}{k_{58}}-g_{58}^{12} \right)=2g_{58}^6k_{58}\sqrt{g_{58}^{12}+g_{58}^{-12}}$$ we get $$\alpha(58) = g_{58}^6k_{58}(99\sqrt{29}-444)$$ as claimed by Borwein brothers in their book Pi and the AGM.

More generally if $p$ is an even positive integer and $n=(p+4)/2$ and there is a formula of type $$nP(q^{2n})-P(q^2)=\frac{4KL}{\pi^2}F_n(l, k) $$ with moduli $k, l$ corresponding to nomes $q, q^n$ then $$P(e^{-2\pi\sqrt{p}}) - \frac{3}{\pi\sqrt{p}}=\left(\frac{2K(k_p)}{\pi}\right)^2\left(\frac{1+k_p^2}{4}+\dfrac{(1-k_p)\Re\left(F_n\left(-k_p, \dfrac{1+k_p} {1-k_p} \right) \right) }{2\sqrt{p}}\right) $$ and we can get a series for $1/\pi$ based on this value as explained below.

Proof of Ramanujan's Fastest Series for $1/\pi$

We start with the formula $$\left(\frac{2K} {\pi} \right)^2=\frac{1}{1+k^2}\sum_{n\geq 0}\frac{(1/4)_n(3/4)_n(1/2)_n} {(n!)^3} \left(\frac{g^{12}+g^{-12}}{2}\right)^{-2n}\tag{28}$$ where $g=(2k/k'^2)^{-1/12}$. Also let us use the notation $$A(k) =\sum_{n\geq 0}\frac{(1/4)_n(3/4)_n(1/2)_n} {(n!)^3} \left(\frac{g^{12}+g^{-12}}{2}\right)^{-2n} $$ and $$B(k) =\sum_{n\geq 0}n\cdot \frac{(1/4)_n(3/4)_n(1/2)_n} {(n!)^3} \left(\frac{g^{12}+g^{-12}}{2}\right)^{-2n} $$ Next we apply the identity $(19)$ on equation $(28) $ to get $$P(q^2)=\frac{g^{12}+(g^{12}-g^{-12})}{2(g^{12}+g^{-12})}A(k)+\frac{3(g^{12}-g^{-12})}{g^{12}+g^{-12}}B(k)\tag{29}$$ Putting $q=e^{-\pi\sqrt{58}}$ so that $$k=k_{58},g=g_{58},K=K(k_{58})$$ we get after some calculation $$P(e^{-2\pi\sqrt{58}})=\frac{99^2+6\cdot 13\cdot 70\sqrt{29}} {4\cdot 99^2} A(k)+\frac{12\cdot 13\cdot 70\sqrt{29}}{2\cdot 99^2}B(k)\tag{30}$$ Using equation $(27)$ the left hand side equals $$\frac{3}{\pi\sqrt{58}}+\left(\frac{1}{4}+\frac{a}{2(1+k^2)\sqrt{58}}\right) A(k) $$ and this simplifies to $$\frac{3}{\pi\sqrt{58}}+\left(\frac{1}{4}+\frac{333}{99\sqrt{29}}\right) A(k) $$ Using this expression in left hand side of $(30)$ we get $$\frac{3} {\pi\sqrt {58}}=\left(\frac{6\cdot 13\cdot 70\sqrt{29}} {4\cdot 99^2}-\frac{333}{99\sqrt {29}}\right) A(k)+\frac{12\cdot 13\cdot 70\sqrt{29}}{2\cdot 99^2}B(k)$$ Multiplying the above equation by $\sqrt{29}/6$ we get $$\frac{1} {2\pi\sqrt {2}}=\frac{13\cdot 70\cdot 29-222\cdot 99} {4\cdot 99^2}\cdot A(k)+\frac{13\cdot 70\cdot 29}{99^2}B(k)\\=\frac{1103A(k)+26390B(k)}{99^2}$$ and thus we get $$\frac{1}{2\pi\sqrt{2}}=\sum_{n\geq 0}\frac{1103+26390n}{99^{4n+2}}\cdot\frac{(1/4)_n(3/4)_n(1/2)_n}{(n!)^3}$$ which is the starting series $(1)$ in symbolic notation.

Note: This post is based on an answer of mine on Mathoverflow which itself is inspired from the work of Pablo F. Refolio.