The following formula for $\pi$ was discovered by Ramanujan: $$\frac1{\pi} = \frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{(4k)!(1103+26390k)}{(k!)^4 396^{4k}}\!$$

Does anyone know how it works, or what the motivation for it is?

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Nick Alger
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    This [article](http://www.google.co.cr/url?sa=t&source=web&cd=20&ved=0CG0QFjAJOAo&url=http%3A%2F%2Fiopscience.iop.org%2F0036-0279%2F62%2F3%2FL17%2Fpdf%2FRMS_62_3_L17.pdf&rct=j&q=ramanujan%27s%20formula%20for%20pi&ei=vngFTZnJOcSblgeRnOzkCQ&usg=AFQjCNFjM_FwCTUinw9A4jJdAEBUzKbkqQ&cad=rja) by W. Zudilin may give you some references where you may look for information about what you ask. It also claims that Ramanujan did not explain how he got his formulas for $1/ \pi$. – Adrián Barquero Dec 13 '10 at 01:46
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    That crazy Ramanujan guy never can tell us anything useful... :D – J. M. ain't a mathematician Dec 13 '10 at 03:08
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    One could build a whole Q&A site out of «what is the motivation for X?» with X running over Ramanujan's identities… – Mariano Suárez-Álvarez May 22 '11 at 15:15
  • I dont think that there is a "motivation" behind this but fun. Ramanujan liked the maths just for fun as for the majority of mathematicians. – Masacroso Sep 17 '14 at 07:24
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    I'm wondering whether all the $99$'s and $4$'s in this are any clue to his method or thought processes. ($9801=99^2$ and $396=4\cdot 99$ do jump out) – timtfj Dec 24 '18 at 00:20
  • @timtfj Unlikely. Ramanujan cited as receiving visions from a certain Goddess (I have since forgotten who). – Frank W Dec 24 '18 at 01:42
  • @timtfj It has to do with the $411$'s, $137$'s; $666$/$616$'s, and $911$'s. Can you find them in the alternate formula, too, of https://en.wikipedia.org/wiki/Chudnovsky_algorithm ? – G. Rem Nov 09 '21 at 22:24
  • Note also the arrangement of the four $4$'s in the former, and, the five $3$'s in the latter ... the numerals with the $k$'s, and $q$'s, respectively, but as dimensions. – G. Rem Nov 09 '21 at 22:35

4 Answers4


Here's an easy introduction to the basics, "Pi Formulas and the Monster Group".


Update: Just to make this more intriguing, define the fundamental unit $U_{29} = \frac{5+\sqrt{29}}{2}$ and fundamental solutions to Pell equations,

$$\big(U_{29}\big)^3=70+13\sqrt{29},\quad \text{thus}\;\;\color{blue}{70}^2-29\cdot\color{blue}{13}^2=-1$$

$$\big(U_{29}\big)^6=9801+1820\sqrt{29},\quad \text{thus}\;\;\color{blue}{9801}^2-29\cdot1820^2=1$$

$$2^6\left(\big(U_{29}\big)^6+\big(U_{29}\big)^{-6}\right)^2 =\color{blue}{396^4}$$

then we can see those integers all over the formula as,

$$\frac{1}{\pi} =\frac{2 \sqrt 2}{\color{blue}{9801}} \sum_{k=0}^\infty \frac{(4k)!}{k!^4} \frac{29\cdot\color{blue}{70\cdot13}\,k+1103}{\color{blue}{(396^4)}^k}$$

See also this MO post.

Tito Piezas III
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This is one of the most interesting results Ramanujan gave and it has a very deep and beautiful theory behind it. Most references regarding this formula try to treat it in high handed manner using modular forms. Ramanujan himself got this formula by remaining within the limits of real analysis and I have presented these ideas along with proofs in my blog post.

Please note that the actual calculation to obtain the numbers 1103 and 26390 in the formula is difficult. Especially no one knows how Ramanujan got 1103 and modern approach to get 1103 is based on numerical calculations.

By Ramanujan's theory (explained in my blog post linked above) we can find infinitely many series of the form $$\frac{1}{\pi} = \sum_{n = 0}^{\infty}(a + bn)d_{n}c^{n}\tag{1}$$ where $a, b, c$ are certain specific algebraic numbers and $d_{n}$ is some sequence of rationals usually expressed in terms of factorials. The modern theory of modular forms allows us to get more details about their algebraic nature (say for example we can get the degree of minimal polynomials of $a, b, c)$. In the case of the current formula it can be shown that both $a, b$ must be quadratic irrationals and $c$ turns out to be a rational number. The calculation of $b, c$ is possible by formulas given by Ramanujan. It is the value of $a$ (related to $1103$) which is difficult to obtain. Now the modern approach goes like this. Since we know the value of $b, c$ and $\pi$ (via some other series calculation) we can find the numerical value of $a$. Knowing that it is a quadratic irrational we can search for integers $p, q, r$ such that $a$ is a root of $px^{2} + qx + r = 0$. This way the quadratic equation is found and the root $a$ is then evaluated in algebraic form.

There are direct formulas to calculate $a, b, c$ and we have two forms of such formulas. One of the forms is a finite formula which may require computations of algebraic nature (so that effectively the value is expressible as a radical expression). Another formula is kind of based on infinite series/product approach which can give numerical values of $a, b, c$. While the algebraic formula for $b, c$ is easy to calculate, the algebraic formula for $a$ is very difficult to compute. Hence the modern approach relies on numerical calculation of $a$. But I very strongly suspect that Ramanujan being an expert in radical manipulation must have found the algebraic value of $a$ using a direct radical manipulation.

In this regard also try to read the book "Pi and the AGM" by Borwein Brothers as they are the first ones to prove this formula of Ramanujan. Also see this answer on mathoverflow for calculation of the constant $1103$.

@Derek Jennings

The general series given in MathWorld is the one discovered by Chudnovsky brothers and it is a different series based on Ramanujan's ideas, but the series in the question under discussion can not be obtained from this general formula of Chudnovsky. A proof of this general series of Chudnovsky is presented in my blog post.

Paramanand Singh
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  • Would the downvoter care to comment? – Paramanand Singh Aug 02 '16 at 20:20
  • A big (+1)! A very thoughtful synopsis on a very interesting and robust topic. – Mark Viola Feb 25 '17 at 18:52
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    Thanks @Dr.MV. I still wonder how Ramanujan developed all of this in pre-independence India (before his association with G H Hardy) with bare minimum resources (not even enough paper to do his work) and his story is one great inspiration for those interested in mathematics. – Paramanand Singh Feb 25 '17 at 20:47
  • Paramanand, you're most welcome. He is one of those phenoms whose talent is difficult to understand. Mozart composed his first symphony at age eight (8) and composed his Symphony #30 at age eighteen (18)! How is that possible? And there are stories of people with Savant Syndrome, who are inexplicably highly developed in areas such as art and music. ... (continued) .... – Mark Viola Feb 25 '17 at 21:37
  • And then, there is [This Case ](http://www.dailymail.co.uk/news/article-2883470/Footballer-woke-coma-car-accident-speaking-fluent-French-thinking-Hollywood-actor-Matthew-McConaughey.html) of a man, who after being struck by lightening, awoke from an induced coma with the ability to fluently speak in French, a language he barely knew and hadn't spoken at all in 12 year. So, how do we explain Ramanujan's unbelievable insights? Perhaps, it is a completely random event of brain configuration - an anomaly that is off the distribution chart. – Mark Viola Feb 25 '17 at 21:37
  • If obtaining $1103$ was any similar to the suggested path to $\pi^4$ (https://mathoverflow.net/questions/232177/the-origin-of-the-ramanujans-pi4-approx-2143-22-identity), we have $$10\pi^4\approx974.0909\approx 974+\frac{9}{99}=\frac{5·2143}{11}$$ $$\frac{99}{2\sqrt{2}\pi}\approx 11.141414\approx 11+\frac{14}{99}=\frac{1103}{99} $$ Maybe the numerical calculations you mention already include this. – Jaume Oliver Lafont May 10 '17 at 09:36
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    @JaumeOliverLafont: the approximation for $\pi^{4}$ is indeed based on numerical values. Borwein brothers in their book *Pi and the AGM* hint that the value $1103$ is also based on numerical evidence. But I doubt this. Ramanujan proved some of his series for $1/\pi$ and gave several details about his method in his paper as well as his Notebooks. And the way Ramanujan handled symbolic and numerical manipulation, I believe he obtained this $1103$ by a direct calculation (although a very tedious calculation). Cont'd... – Paramanand Singh May 10 '17 at 10:29
  • @JaumeOliverLafont: I think the modern development of modular forms and other related stuff has pushed mathematicians in a direction which is very orthogonal to Ramanujan's methods and thus people are not able to discern his methods properly. – Paramanand Singh May 10 '17 at 10:32

Here is a nice article Entitled: "Ramanujan's Series for $\displaystyle\frac{1}{\pi}$ : A Survey", by Bruce C.Berndt. This article appeared in the American Mathematical Monthly *August/September* 2009. You can see it here.

  • @Nick:Hopefully this will help. –  May 22 '11 at 11:57
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    You are saying he is the only person you know personally that is working on the subject, or that he is the only person working on the subject? One of these two claims is of limited interest and the other is false :) – Mariano Suárez-Álvarez May 22 '11 at 15:18
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    @Mariano: Then why are you interested. Well, I know he works on Ramanujan related mathematics. I have not heard anyone else name other than him. –  May 22 '11 at 15:24
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    Well, I was interested because, if you actually thought that he is the only person working on the subject, then you might be interested in correcting that impression, and I as well as others might help you with that. You have managed to uninterest me, though! – Mariano Suárez-Álvarez May 23 '11 at 01:51
  • @Mariano: Sorry Mariano. If I managed to do like that. –  May 23 '11 at 04:51
  • Wow thanks for that link Chandru. – Nick Alger Jun 03 '11 at 03:47
  • @Nick: Welcome :) –  Jun 03 '11 at 04:47

The explanation for the existence of this series is given here. Search for the phrase "The general form of the series is" to locate it. The series cited in the question appears immediately before the explanation.

Derek Jennings
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    Derek, thanks for the link. However I was hoping for a bit more indepth of an explanation than the short comment that is written there ("Equation (78) is derived from a modular identity of order 58"). – Nick Alger Dec 14 '10 at 02:03