On Ramanujan's fastest series.

Question

Context

With some effort we can show that Ramanujan's fastest series implies: \begin{align} \frac{8E(k_{58})K(k_{58})}{\pi^2}-\frac{aK^2(k_{58})}{\pi^2}=\frac{\sqrt{58}}{29\pi},\tag{1} \end{align} with: $a=\left(\frac{489227532 \sqrt{58}}{{29}}- \frac{691872192\sqrt{29}}{29}+90847272\sqrt{2}-128477440\right)$ and $k_{58}=(13\sqrt{58}-99)(\sqrt{2}-1)^6.$

Some related formulas

Being: \begin{align} P(q)=1-\sum_{n=1}^{\infty}\frac{24nq^{2n}}{1-q^{2n}},\tag{2} \end{align} with $q=e^{-\pi\frac{K(k')}{K(k)}}$, $K(k)=\int_{0}^{\pi/2}\frac{dt}{\sqrt{1-k^2\sin^2t}}$ and $E(k)=\int_{0}^{\pi/2}{\sqrt{1-k^2\sin^2t}}dt$ the complete elliptic integrals of the first kind and second kind respectively and $k'=\sqrt{1-k^2}$ the complementary modulus. Then we have: \begin{align} P(q)=\frac{12E(k)K(k)}{\pi^2}+\frac{(4k^2-8)K^2(k)}{\pi^2}\tag{3}. \end{align} Ramanujan dealt with $P(q)$ proving (as Paramanand has noted) that $P(q)$ would have the form: \begin{align} P(q)=\frac{K^2(k)}{\pi^2}A(n,k)+\frac{3}{\pi\sqrt{n}},\tag{4} \end{align} where $A(n,k)$ is an algebraic number. Asumming this we have for $n=58$: \begin{align} \frac{8E(k_{58})K(k_{58})}{\pi^2}+\frac{2(4k_{58}^2-8)K^2(k_{58})}{3\pi^2}=\frac{K^2(k_{58})}{\pi^2}A(58,k_{58})+\frac{\sqrt{58}}{29\pi}. \end{align} Comparing this last with $(1)$ gives $A(58,k_{58})$. This suggests that some reverse approach can be done to obtain the modular equation of degree 29 and the $f(k)$ in the following formula: \begin{align} 29P(q^{29})-P(q)=\frac{K^2(k)}{\pi^2}f(k). \end{align}

Note 1

ASAP we are going to cover the case $k_{163}$ (Chudnovsky's formula), since there is a form like $(1)$ also for this case. We believe that a hand proof exists for both series.

1. See this answer of mine dealing with $k_{1}$ in an elementary approach: Finding $p\cdot K\left(s\right)E\left(s\right)+q\cdot K\left(s\right)^{2}$ such that it is a known constant
2. And this answer of @ParamanandSingh dealing with $k_{5}$ using Ramanujan ideas: A closed form for a triple integral involving Heron's formula

Note 2: Updated 30/03/2024

We have: \begin{align} \frac{4K^2(k)}{\pi^2}=\frac{1}{k^2+1}\sum_{n=0}^{\infty}k^{2n}\left( \frac{k^2-1}{4(k^2+1)^2}\right)^{2n}\frac{(4n)!}{n!^4}\tag{5} \end{align} and \begin{align} \frac{8E(k)K(k)}{\pi^2}=\sum_{n=0}^{\infty}\frac{2k^{2n}(k^2-1)^{2n}(k^2+1)^{-2(2n+1)}\left(k^4n-k^2(6n+1)+n+1 \right)(4n)!}{2^{4n}n!^4}\tag{6} \end{align} The first formula is the same as given in the below answer. We left to the reader the proof of the second formula. Since $K(k'_{58})=\sqrt{58}K(k_{58})$ then using Legendre's relation, $(5)$ and $(6)$ with some effort one can get with full rigour: \begin{align} \frac{1}{2\sqrt{58}\pi}=\frac{K(k_{58})E(k'_{58})}{\sqrt{58}\pi^2}+\frac{455\sqrt{29}}{9801}\sum_{n=0}^{\infty}\frac{n(4n)!}{396^{4n}n!^4}+\left(\frac{455\sqrt{29}}{39204}-\frac{1}{16} \right)\sum_{n=0}^{\infty}\frac{(4n)!}{396^{4n}n!^4}.\tag{7} \end{align} A possible approach is to show that: \begin{align} \frac{K(k_{58})E(k'_{58})}{\sqrt{58}\pi^2}=\frac{\sqrt{58}}{232\pi}+\left(\frac{1}{16}-\frac{37\sqrt{29}}{3828} \right)\sum_{n=0}^{\infty}\frac{(4n)!}{396^{4n}n!^4}\tag{8} \end{align} This form is better than $(1)$ because a lot of simplification occurs. Then combining $(7)$ and $(8)$ we have: \begin{align} \frac{1}{2\sqrt{58}\pi}=\frac{455\sqrt{29}}{9801}\sum_{n=0}^{\infty}\frac{n(4n)!}{396^{4n}n!^4}+\frac{1103\sqrt{29}}{568458}\sum_{n=0}^{\infty}\frac{(4n)!}{396^{4n}n!^4}+\frac{\sqrt{58}}{232\pi}.\tag{9} \end{align} which is Ramanujan's fastest series.

Question

Can we prove $(1)$ or $(8)$ by hand like Ramanujan did?

Answer 1

The short answer to your question is that there is no available proof for the identity $(1)$ of your question which involves only hand calculation. In the following I present reasonable details of work by Ramanujan as well Borwein brothers.

In his famous paper Modular equations and approximations to $\pi$ Ramanujan discussed about evaluation of the function $P(q^2) $ where $$P(q) =1-24\sum_{n\geq 1}\frac{nq^n}{1-q^n}\tag{1}$$ for certain specific values of $q$ of the form $q=\exp(-\pi\sqrt{r}) $ where $r\in\mathbb{Q},r>0$. The function $P(q) $ is related to Dedekind's eta function $$\eta(q) =q^{1/24}\prod_{n\geq 1}(1-q^n),\tag{2}$$ via $$P(q) = 24q\frac{d}{dq}\log\eta(q)\tag{3}$$

The topic is highly interesting and is related to theory of elliptic integrals which we discuss in brief. Let us start with definitions of elliptic integrals first. Let $k\in(0,1)$ be the elliptic modulus and $k'=\sqrt{1-k^2}$ be the complementary modulus and we define $$K(k)=\int_0^{\pi/2}\frac{dx}{\sqrt{1-k^2\sin^2x}},E(k)=\int_0^{\pi/2}\sqrt{1-k^2\sin^2x}\,dx\tag{4}$$ If values of $k, k'$ are available from context we write $K, K', E, E'$ for $K(k), K(k'), E(k), E(k') $ respectively. These integrals satisfy a relation $$KE'+K'E-KK'=\frac{\pi} {2}\tag{6}$$ which is famous by the name Legendre's identity. The identity however fails to give a direct link between $K, E$ and this makes the evaluation of $E$ considerably difficult compared to that of $K$. Ramanujan gave a very ingenious technique of evaluation of $E$ in his aforementioned paper via the use of function $P(q^2)$. If the values of $K, K'$ are known then the values of $k, k'$ can be obtained using theta functions of Jacobi as $$k=\frac{\vartheta_2^2(q)}{\vartheta_3^2(q)},k'=\frac{\vartheta_4^2(q)}{\vartheta_3^2(q)}\tag{7}$$ where \begin{align} q&=\exp(-\pi K'/K) \tag{8a}\\ \vartheta_2(q)&=\sum_{n\in\mathbb{Z}}q^{(n+(1/2))^2}\tag{8b}\\ \vartheta_3(q)&=\sum_{n\in\mathbb{Z}}q^{n^2}\tag{8c}\\ \vartheta_4(q)&=\vartheta_3(-q)\tag{8d} \end{align} The variable $q$ is called the nome corresponding to modulus $k$ and we also have another relation $$\vartheta_3^2(q)=\frac{2K}{\pi}\tag{9}$$ Now we talk about the relation between $k, q$ given by $q=e^{-\pi K'/K} $. It can be proved that $K'/K$ is a decreasing function of $k$ and maps $(0,1)$ to $(0,\infty)$ and hence $q$ is an increasing function of $k$ and maps $(0,1)$ to itself. More precisely we have $$\frac{dq} {dk} =\frac{\pi^2q}{2kk'^2K^2}\tag{10}$$ Using product representations of theta functions it can be proved that $$\eta(q^2)=2^{-1/3}\sqrt{\frac{2K}{\pi}}(kk')^{1/6}\tag{11}$$ and using $(3),(10),(11)$ we can prove that $$P(q^2) =\left(\frac{2K}{\pi}\right)^2\left(\frac{3E}{K}+k^2-2\right)\tag{12}$$ so that the value $E$ can be computed if $K, P(q^2)$ are known.

We now describe major theorems of Jacobi, Ramanujan and Selberg-Chowla which help us to perform these evaluations (at least in theory). Let $r$ be a positive real number and let $k$ be a given modulus then the value $rK'/K$ is positive and hence there is a unique modulus $l\in(0,1)$ such that $K(l') /K(l) =rK'/K$ so that $l$ is a function of $k$ if $r$ is a positive constant. The integrals $K(l), K(l') $ are traditionally denoted by $L, L'$ and Jacobi proved

Theorem 1 (Jacobi): If $r$ is a positive rational number and $L'/L=rK/K$ then the relation between $k, l$ is algebraic ie there is a polynomial $P(x, y) $ with integer coefficients such that $P(k, l) =0$. This relation between $k, l$ is said to be a modular equation of degree $r$.

Ramanujan added a further constraint in this equation namely $l=k'$ so that $P(k, k') =0$ and then both $k, k'$ are algebraic numbers and $L'=K, L=K'$ so that $K'/K=1/\sqrt{r}$ and $L'/L=\sqrt {r} $. And then we have

Theorem 2: If $r$ is a positive rational number and $q=\exp(-\pi\sqrt{r}) $ then the values $k, k'$ given in $(7)$ are algebraic and are called singular moduli and denoted by $k_r, k'_r$.

Ramanujan was an expert in finding modular equations and evaluating singular moduli in explicit radical form. Ramanujan further proved

Theorem 3 (Ramanujan): If $r$ is a positive rational number and moduli $k, l$ are connected by modular equation of degree $r$ so that $L'/L=rK'/K$ then $$rP(q^{2r})-P(q^2)=\frac{4KL}{\pi}A_r(l,k)\tag{13}$$ where $A_r(l, k) $ is an algebraic function of $l, k$.

We also have a related formula (which is a consequence of the transformation formula for Dedekind's eta function) $$rP(e^{-2\pi\sqrt{r}})+P(e^{-2\pi/\sqrt{r}})=\frac{6\sqrt{r}}{\pi}\tag{14}$$ which holds for all positive real numbers $r$. Putting $q=e^{-\pi/\sqrt{r}} $ in $(13)$ and using $(14)$ Ramanujan proved

Throrem 4 (Ramanujan): If $r$ is a positive rational number and $q=e^{-\pi\sqrt{r}} $ then $$P(q^2)=\frac{3}{\pi\sqrt{r}}+A_r\cdot\left(\frac{2K}{\pi}\right)^2\tag{15}$$ where $A_r$ is an algebraic number dependent on $r$.

Ramanujan gave formulas (ie explicit radical expressions) for $A_r(l, k) $ in terms of $k, l$ for many positive integer values of $l$. He mentioned that such a formula can be obtained by differentiating the modular equation $P(k,l)=0$ twice with respect to $k$. The technique works fine for small values of $r=2,3$ but becomes unwieldy for larger values and no one really knows how Ramanujan got those not so unwieldy expressions for $A_r(l, k) $. Perhaps his great and unmatched skill in the manipulation of radicals helped him here. Using these expressions for $A_r(l, k) $ and the values of singular moduli $k_r, k'_r$ one can get the values of the constant $A_r$ in $(15)$.

Using $(12)$ and $(15)$ we can prove that $$K(k_r) E(k_r) + B_rK(k_r) ^2=\frac{\pi}{4\sqrt{r}}\tag{16}$$ where $r$ is a positive rational number and $B_r$ is some algebraic number dependent on $r$. This is exactly what your equation $(1)$ in question says for $r=58$. But to evaluate the constants $A_r, B_r$ it is essential to know the explicit form of function $A_r(l, k) $ and the values of related singular moduli. These constants $A_r, B_r$ are also needed to find certain series for $1/\pi$ which were first brought into prominence by Ramanujan. Before discussing this we also mention a theorem by Selberg-Chowla which deals with evaluation of $K(k_r) $.

Theorem 5 (Selberg-Chowla): If $r$ is a positive rational number then the value $K(k_r) $ can be expressed in a closed form involving $\pi$ and values of Euler's Gamma function at rational points. There is a direct and nice formula available if $r$ is a square free positive integer and $\mathbb{Q} (\sqrt{-r}) $ has class number $1$.

Using these aforementioned theorems it is possible in principle to evaluate $K(k_r), E(k_r) $ in closed form consisting of values of Gamma function at rational points. But in practice the biggest challenge is to find the constants $A_r, B_r$.

Borwein brothers approach this problem in a slightly different manner and they define a function $\alpha:(0, \infty) \to\mathbb{R} $ via $$\alpha(r) =\frac{E'} {K} - \frac {\pi} {4K^2}\tag{17}$$ with modulus $k$ corresponding to nome $q=e^{-\pi\sqrt{r}} $. Using Legendre's identity and the fact that $K'=\sqrt{r} K$ we can rewrite $(17)$ as $$\frac{\pi} {4}=\sqrt{r} KE-(\sqrt{r} - \alpha(r)) K^2\tag{18}$$ and from $(16)$ we see that $$B_r=\frac{\alpha(r)} {\sqrt{r}} - 1$$ so that $\alpha(r) $ is algebraic when $r$ is rational.

Borwein brothers also give an expression for $\alpha(r) $ in terms of theta functions which can be used for numerical evaluation $$\alpha(r) =\frac{1}{\vartheta_3^4(q)}\left(\frac{1}{\pi}-4q\sqrt{r}\frac{d}{dq}\log\vartheta_4(q)\right), q=e^{-\pi\sqrt{r}} \tag{19}$$ and show that $\alpha (r) \to1/\pi$ as $r\to\infty$.

In their book Pi and the AGM Borwein brothers give a host of formulas to evaluate $\alpha(r) $ and also indicate its role in deriving Ramanujan type series for $1 /\pi$. I will mention here a few such formulas below. Let us use the symbol $k_r$ not just for singular modulus (for rational $r$) but for the modulus corresponding to nome $q=e^{-\pi\sqrt{r}} $ and then we have $$\alpha(4r)=(1+k_{4r})^2\alpha(r)-2\sqrt{r}k_{4r}\tag{20}$$ and $$\alpha(9r)=s(r)^2\alpha(r)-\frac{\sqrt{r}(s(r)^2+2s(r)-3)} {2} \tag{21}$$ where $$s(r) =\sqrt{1+4\cdot\frac{(k_{9r}k'_{9r})^{3/4}}{(k_rk'_r)^{1/4}}}\tag{22}$$ There are formulas available for $\alpha(2r),\alpha(3r)$ as well but they are a bit complicated. The basis of these formulas is the expression $A_r(l, k) $ which was given explicitly as radicals by Ramanujan for many integer values of $r$ and Borweins say that these radical expressions can be verified for small values of $r=2,3,4$ and for higher values we rely on Ramanujan. These expressions were later verified using theory of modular forms and mathematical software by Bruce C. Berndt and his collaborators.

Let us now discuss how Borweins established the Ramanujan type series for $1/\pi$. They start with a formula of the form $$\left(\frac{2K}{\pi}\right)^2=m(k) \sum_{n\geq 0}a_n\phi(k)^n=m(k) F(\phi(k)) \tag{23}$$ where $m(k), \phi(k) $ are algebraic functions of $k$ and the above series is a hypergeometric type series. Differentiating the series with respect to $k$ we get $$\frac{8K}{\pi^2}\frac{dK}{dk}=m'(k)\sum_{n\geq 0}a_n\phi(k) ^n+m(k)\sum_{n\geq 1}na_n\phi(k) ^{n} \frac{\phi'(k)} {\phi(k)}\tag{24} $$ Now we can note that $$\frac{dK} {dk} =\frac{E-k'^2K}{kk'^2}$$ and using $(18)$ to express $E$ terms of $K$ we get $$\frac{1}{\pi}=\sqrt{r}k_r{k'}_r^2\cdot\frac{4K}{\pi}\frac{dK}{dk}+(\alpha(r) - \sqrt{r} k_r^2)\frac{4K^2}{\pi^2}$$ Using $(23),(24)$ in above equation we get the following series for $1/\pi$ $$\frac{1}{\pi}=\sum_{n\geq 0}a_n\left(\frac{\sqrt{r}} {2}\cdot kk'^2m'(k)+(\alpha(r) - \sqrt{r} k^2)m(k) +\frac{n\sqrt{r}} {2}\cdot kk'^2m(k)\cdot\frac{\phi'(k)}{\phi(k)}\right) \phi(k) ^n\tag{25}$$ with $k=k_r$.

Ramanujan's fastest series is based on the formula $$\frac{4K^2}{\pi^2}=\frac{1}{1+k^2}{}_3F_2\left(\frac{1}{4},\frac{3}{4},\frac{1}{2};1,1;\left(\frac{g(k)^{12}+g(k)^{-12}}{2}\right)^{-2}\right)$$ where $g(k) =(2k/k'^2)^{-1/12}$. Using $m(k) =(1+k^2)^{-1}$ and $\phi(k) =4/(g(k)^{12}+g(k)^{-12})^2$ in $(25)$ one gets the following series $$\frac{1}{\pi}=\sum_{n\geq 0}\frac{(1/4)_n(1/2)_n(3/4)_n}{(n!)^3}d_n(r)x_r^{2n+1}\tag{26}$$ where $$x_r=\frac{4k_r{k'}_r^2}{(1+k_r^2)^2}\tag{27}$$ and $$d_n(r) =\left(\frac{\alpha(r) /x_r} {1+k_r^2}-\frac{\sqrt{r}}{4}g(k_r)^{-12}\right) +n\sqrt{r} \left(\frac{g(k_r) ^{12}-g(k_r)^{-12}}{2}\right)\tag{28} $$ Ramanujan's series in question is obtained by putting $r=58$ in above formula. The stumbling block however is the calculation of $\alpha(r)$ for $r=58$ and Borweins mention that they first evaluated $d_0(58)$ given in $(28)$ numerically and then compared the first term of their series with that given by Ramanujan (containing the constant $1103$) and argued that if their difference were beyond a certain margin of error the series would differ from $1/\pi$ by a different margin. That the series sums to $1/\pi$ within a certain margin of error confirms that $d_0(58)$ must match corresponding value of the first term as given by Ramanujan. And then they evaluated $\alpha(58)$ in symbolic form. A similar procedure was used to evaluate $\alpha(37)$ based on a different series given by Ramanujan.

To the best of my knowledge (from online sources including math overflow and mathse) there is no hand calculation available for $\alpha(37),\alpha(58),\alpha(163)$ and most of the modern work in this direction is highly assisted by mathematical software.