How to prove
$$ \frac{\sqrt{8}}{9801}\sum_{n=0}^{\infty}\frac{(4n)!(1103+26390n)}{(n!)^4 396^{4n}}=\frac{1}{\pi}, $$
which actually looks like coincidence?
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Nothing is a coincidence in math – mathworker21 Dec 23 '18 at 22:41
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@mathworker21 Not even $\pi^4+\pi^5≈e^6$? (It's ridiculously close.) – timtfj Dec 23 '18 at 23:52
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@timtfj you never know. here is an explanation for why $e^{\pi\sqrt{163}}$ is very close to an integer. http://people.maths.ox.ac.uk/greenbj/papers/ramanujanconstant.pdf – mathworker21 Dec 24 '18 at 00:50
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1@mathworker21 The first sentence is somewhat startling in itself. "Ramanujan observed that [insane calculation]" – timtfj Dec 24 '18 at 01:08
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If you want the proof then you have to understand that it's really difficult. The conceptual ideas can be grasped with some effort but evaluation of the constant $1103$ in your formula is very difficult. – Paramanand Singh Dec 25 '18 at 08:26