Ramanujan's famous pi formula states that \begin{equation} \frac{1}{\pi}=\frac{2\sqrt{2}}{99^2}\sum_{k=0}^{\infty}\frac{(4k)!}{k!^4}\frac{26390k+1103}{396^{4k}} \end{equation} How can one prove this? If the proof is too long for this site, you can reference any article containing the proof.

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    $\begingroup$ I added some tags. I also found, after searching for "Ramanujan Sato series proof" an AMM article Ramanujan's Series for 1/π: A Survey doi.org/10.1080/00029890.2009.11920975 (pdf), which references J. M. Borwein and P. B. Borwein, Pi and the AGM; A Study in Analytic Number Theory and Computational Complexity, Wiley, New York, 1987 (publisher page. as containing the first proofs. $\endgroup$ – David Roberts Oct 10 at 7:19
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    $\begingroup$ This should also help: en.wikipedia.org/wiki/Ramanujan%E2%80%93Sato_series $\endgroup$ – David Roberts Oct 10 at 7:20
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    $\begingroup$ Borwein, J. M.; Borwein, P. B.; Bailey, D. H. (1989). "Ramanujan, modular equations, and approximations to pi; Or how to compute one billion digits of pi" (PDF). Amer. Math. Monthly. 96 (3): 201–219. doi.org/10.1080%2F00029890.1989.11972169 might be better. $\endgroup$ – David Roberts Oct 10 at 9:58
  • $\begingroup$ All the articles were good. $\endgroup$ – mathguy Oct 10 at 12:16

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