# 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]

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

Ramanujan is the BEST.

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