The formula $$\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n+41}{25920^n}=\frac{288\sqrt{5}}{5\pi}$$ (in older notation) appears as eq. 38 in Ramanujan's paper Modular equations and approximations to $\pi$; $(a)_n$ is the Pochhammer symbol.

But – the formula is like an exercise for the reader. Allegedly, it can be deduced from the theory of modular equations, but how exactly?

It should be somehow possible to prove the formula from Clausen's formulas and representations of $$25P(q^{50})-P(q^2)$$ where $$P(q)=1-24\sum_{k=1}^\infty \frac{kq^k}{1-q^k}$$ with $|q|\lt 1$. The hypergeometric representation of Ramanujan's formula is known to be $$41\,_3F_2\left(\frac14,\frac12,\frac34,1,1,-\frac{1}{25920}\right)-\frac{161}{69120}\, _3F_2\left(\frac54,\frac32,\frac74,2,2,-\frac{1}{25920}\right)=\frac{288\sqrt{5}}{5\pi}.$$

What I tried

Let $$K(x)=\int_0^{\pi/2}\dfrac{dt}{\sqrt{1-x\sin^2 t}}$$ and $$E(x)=\int_0^{\pi/2}\sqrt{1-x\sin^2 t}\,dt$$ be the elliptic integrals. Using the hypergeometric differential equation, the problem at hand shoud then be reducible to $$K\left(\dfrac{1}{2}-6\sqrt{-360+161\sqrt{5}}\right)\left(2E\left(\dfrac{1}{2}-6\sqrt{-360+161\sqrt{5}}\right)-\left(1+\sqrt{-840+376\sqrt{5}}\right)K\left(\dfrac{1}{2}-6\sqrt{-360+161\sqrt{5}}\right)\right)=\frac{\pi}{10}.$$

Now I recognized the argument of the elliptic integrals as a special value of the modular lambda function $\lambda$ (https://en.wikipedia.org/wiki/Modular_lambda_function): $$\lambda (5i)=\dfrac{1}{2}-6\sqrt{-360+161\sqrt{5}}.$$

So if someone can prove that $$K(\lambda (5i))=\dfrac{\sqrt{5}+2}{20}\dfrac{\Gamma (1/4)^2}{\sqrt{\pi}}$$ and $$E(\lambda (5i))=\dfrac{(-2+\sqrt{5})\pi^{3/2}}{\Gamma (1/4)^2}+\dfrac{\left(2+\sqrt{5}+2\sqrt{-10+6\sqrt{5}}\right)\Gamma (1/4)^2}{40\sqrt{\pi}},$$ (that's out of my reach), then Ramanujan's formula is proved.


I noticed that $K(\lambda (5i))$ can be easily proved to have the desired closed form by using division values of elliptic functions with parameter $-1$. But evaluating $E(\lambda (5i))$ seems hard...

This question was also asked on MSE: https://math.stackexchange.com/questions/4898590/how-did-ramanujan-find-sum-n-0-infty-1n-frac1-2-n1-4-n3-4-nn


2 Answers 2


There are (at least) two ways to prove Ramanujan-like formulas for $1/\pi$: either using elliptic functions and modular equations, as you try above, or by using modular functions, which is not the same. This second approach is usually much simpler and more systematic. Everything (and much more), including the proof of the formula you quote is in my joint paper with J. Guillera arXiv:2101.12592v1

  • $\begingroup$ How do I evaluate $E_2\left(\frac{1+i\sqrt{25}}{2}\right)$? Isn't that equally hard as evaluating $E(\lambda (5i))$ (as in the end of my post)? (sorry, I'm new to this) $\endgroup$
    – Nomas2
    Commented Apr 15 at 21:59
  • $\begingroup$ I know that Chowla–Selberg gives $\eta$ which gives $E_4$ and $E_6$, but how should I evaluate $E_2$? $\endgroup$
    – Nomas2
    Commented Apr 17 at 8:58
  • 1
    $\begingroup$ See my answer to your new post. $\endgroup$ Commented Apr 17 at 15:10

The formula in the OP is a hypergeometric function identity, $$\frac{1}{41}\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n+41}{25920^n}=$$ $$\qquad= \,_4F_3\left(\frac{1}{2},\frac{1}{4},\frac{3}{4},\frac{685}{644};1,1,\frac{41}{644};-\frac{1}{25920}\right)=\frac{288 \sqrt{5}}{205 \pi }.$$ This is equation (35), given with a proof by M. I. Qureshi in On Certain Hypergeometric Summation Theorems Motivated by the Works of Ramanujan, Chudnovsky and Borwein

  • $\begingroup$ The linked paper of Qureshi just rewrites the formulas of Ramanujan, Borwein, Chudnovsky in terms of the hypergeometric ${}_4F_{3}$ function and this is nothing more than notational convenience. No proofs of any of the formulas are contained in that paper. The paper should be treated with a high degree of skepticism. $\endgroup$ Commented Apr 26 at 10:25

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.