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} $

Question

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.

Answer 1

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