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.

  • $\begingroup$ See cecm.sfu.ca/~pborwein/PAPERS/P40.pdf. $\endgroup$ Apr 20, 2014 at 7:55
  • $\begingroup$ @DietrichBurde: I read that paper and found that regarding this series Borwein mentions about calculation of $\alpha(58)$ which leads to $1103$ in the series. He writes "It is less clear how one calculates $\alpha(58)$ in algebraic form... but a numerical calculation ...is easily obtained" and this confirms the value $1103$. Borwein believes that "this is presumably what Ramanujan observed". Then from the number $1103$ and value of $g_{58}$ the value of $\alpha(58)$ in algebraic form is obtained. $\endgroup$ Apr 20, 2014 at 8:12
  • $\begingroup$ @DietrichBurde: Continuing from prev comment. Borwein then says that the agreement of the sum of series with $1/\pi$ to 3 millions places confirms that $\alpha(58)$ should have a specific algebraic value. It then seems that no one has really calculated the value of $\alpha(58)$ directly by manipulating radicals/solving equation etc, but all is based on numerical calculations and then some knowledge about algebraic nature of $\alpha(58)$. I strongly believe that Ramanujan did calculate the value $1103$ exactly and not a guess based on numerical calculations. $\endgroup$ Apr 20, 2014 at 8:15

4 Answers 4


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








2. Following H. M. Weber, one has


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


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.

  • 1
    $\begingroup$ This is much beyond my understanding and I need to study a lot more stuff to figure all this out in detail. Also I am not able to understand the definition of $[a, b, c, d]$ because the defining equation does not have $a, b, c, d$ on right side. Also $\delta$ is supposed to be a positive divisor of $N = 58$, but no idea as to what is $a_{\delta}$. Some more information on this would definitely help. +1 by the way to get the desired value of the constant $G_{0}$. $\endgroup$ Mar 12, 2015 at 4:10
  • 1
    $\begingroup$ @ParamanandSingh: I have corrected some typos and rewritten the definition of $[\cdots, a_\delta, \cdots]$. More details would be added in the near future. $\endgroup$
    – Y. Zhao
    Mar 12, 2015 at 9:09

In https://arxiv.org/abs/1911.03968, I provide a complete proof of the Ramanujan's series for $1/\pi$ in the question. First, we obtain, using a Maple procedure, a modular equation of level $2$ and degree $29$ in the Russell form: $$ u^2=\alpha \beta, \quad v^2=(1-\alpha)(1-\beta), \quad P(u,v)=0, $$ where $P(u,v)$ is a long symmetric polynomial of degree $15$ in both variables $u$ and $v$ (For readers that do not use Maple, we show the modular equation in the Appendix). From it, we get the proof of that impresive series for $1/\pi$ due to Ramanujan.

  • 1
    $\begingroup$ I will have a look at your paper. Thanks and +1. Is there no way to do it by hand? Perhaps the way Ramanujan would have done? The great misfortune of Ramanujan was his poverty. He didn't record his derivations primary due to lack of paper. $\endgroup$ Nov 12, 2019 at 2:29
  • $\begingroup$ Modular equations are not unique. Perhaps it is possible to find a simpler one. The computations in the paper are easy by computer, but I cannot imagine how Ramanujan could have got the number 1103 (if he really got it rigorously). On the other hand and curiously, for obtaining this number, I have used a formula for the multiplier which is due to Ramanujan. $\endgroup$ Nov 12, 2019 at 2:46
  • $\begingroup$ Ramanujan did give a modular equation for degree 29 but it was same as the one by Russell (Bruce Berndt mentioned in one of his Ramanujan Notebooks), but I think the same modular equation could havee been simpler if presented in the alternative theory for $s=1/4$ dealing with $F(1/4,3/4;1;\alpha)$. This is what you have tried to do with polynomial equation in $u, v $. I wonder if Ramanujan probably derived another equation in the form $m=f(\alpha, \beta) $ for degree 29. $\endgroup$ Nov 12, 2019 at 3:40

See: J.M Borwein, P.B. Borwein, "Pi and AGM". John Wiley and Sons, Inc. New York, Chichester, Brisbane, Toronto, Singapore, 1987.

pages 172, 177.

I have a note and a Berndt's formula

1) Let $$ A_{p,r}:=\frac{f(-q^2)}{q^cf(-q^{2p})}\textrm{, }c=\frac{p-1}{12} $$ then $T_{p,r}=P(q)-pP(q^p)$ is $$ T_{p,r}=\frac{24\sqrt{r}}{\pi A_{p,r}}\frac{dA_{p,r}}{dr}. $$ 2) The formula in Berndt's book [1] is $$ T_{p,r}=\frac{16}{\pi^2}(k_rk'_r)^2K^2\frac{d}{dk}\log\left(m_{p}(r)^{3}\frac{k_rk'_r}{k_{p^2r}k'_{p^2r}}\right) $$

[1]: B.C. Berndt. 'Ramanujan's Notedbooks Part III'. 1991 ed., New York: Springer-Verlag.

  • $\begingroup$ I have that book and unfortunately it does not provide the details of the calculation for this specific series of Ramanujan. $\endgroup$ Mar 12, 2016 at 13:27
  • $\begingroup$ Have you try to use the formula $\endgroup$ Oct 28, 2016 at 20:44
  • $\begingroup$ I had tried to obtain the expression for $m$ by differentiating the modular equation of degree 29. But this lead to very complex expressions and again differentiating it was simply not possible for me. Is there an expression for multiplier $m$ for degree 29 available in literature? $\endgroup$ Oct 29, 2016 at 4:01

I analyzed the series in question as well as the one given by Chudnovsky brothers and also the following series given by Ramanujan $$\frac{4}{\pi}=\sum_{n=0}^{\infty} (-1)^n\frac{(1/4)_n(2/4)_n(3/4)_n}{(n!)^3}(1123+21460n)\left(\frac{1}{882}\right)^{2n+1}$$ All these series which can be obtained from Ramanujan's technique can be expressed in the format $$\frac{1}{\pi}=\sum_{n=0}^{\infty} (a_N+nb_N)d_n (c_N) ^{n} $$ where $N$ is a positive integer and $a_N, b_N, c_N $ are algebraic numbers dependent on $N$ and are based on modular equations of degree $N$. Further $d_n$ is a typical sequence involving ratio of rising factorials.

In each case it can be proved that $a_N\sqrt{N} /b_N\to1/\pi$ as $N\to\infty $. This is quite obvious if one studies the formulation given by Chudnovsky brothers. For the technique provided by Ramanujan it needs some patience to verify this. This is also related to the fact that the function $\alpha(N) $ given by Borwein brothers in Pi and the AGM tends to $1/\pi$.

By a more careful analysis of the rate of convergence of the ratio $a_N\sqrt {N} /b_N$ it is possible to evaluate $a_N$ given the value $b_N$.

One can easily find a suitable rational approximation $p/q$ for $1/\pi\sqrt {N} $ and then just use $a_N=pb_N/q$.

This is so evident in the approximations $$\frac{1123}{21460}\approx\frac{1}{\pi\sqrt{37}}$$ and $$\frac{1103}{26390}\approx\frac{1}{\pi\sqrt{58}}$$ and $$\frac{13591409}{545140134} \approx\frac{1}{\pi\sqrt{163}}$$ Even for small values of $N$ eg $N=7$ the following approximation $1/\pi\sqrt{7}\approx 5/42$ works fine to give the series $$\frac{1}{\pi}=\sum_{n=0}^{\infty} \binom{2n}{n}^3\frac{42n+5}{2^{12n+4}}$$ (no other integer $p$ except $5$ gives a better approximation of the form $p/42$ for $1/\pi\sqrt{7}$).


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