# What do I need to know to prove this? $\frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{ (4k)! (1103+26390k) }{ (k!)^4 396^{4k} } = \frac1{\pi}$ [duplicate]

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This is an identity put forward by Ramanujan (often used as "proof" of his genius):

$$\frac{2\sqrt{2}}{9801} \sum_{k=0}^\infty \frac{ (4k)! (1103+26390k) }{ (k!)^4 396^{4k} } = \frac1{\pi}$$

How does one go about proving this? Alternatively, what does one need to know to be able to do so?

Any help is appreciated.

## marked as duplicate by 6005, Mark Bennet, Davide Giraudo, Vedran Šego, dtldarekSep 27 '13 at 20:36

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

## 1 Answer

Ramanujan is the BEST.

Visit this for an insight, well this insight is far more deeper than put

http://en.wikipedia.org/wiki/Srinivasa_Ramanujan

https://sites.google.com/site/tpiezas/0013

http://mathworld.wolfram.com/PiFormulas.html

http://paramanands.blogspot.in/2012/03/modular-equations-and-approximations-to-pi-part-1.html#.UkXdl9Kl55A

• Thanks! The links are really good, and since this has been marked as a dup I'll accept your answer. – Soham Chowdhury Sep 28 '13 at 2:38
• @SohamChowdhury Hail Ramanujan – Shobhit Sep 28 '13 at 3:24