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2-2-9801-n-1-4n-1103-26390n-n-4-396-4n-1-pi-Prove-that-




Question Number 117460 by Dwaipayan Shikari last updated on 11/Oct/20
((2(√2))/(9801))Σ_(n=1) ^∞ (((4n)!(1103+26390n))/((n!)^4 396^(4n) ))=(1/π)   (Prove that)
$$\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{9801}}\underset{{n}=\mathrm{1}} {\overset{\infty} {\sum}}\frac{\left(\mathrm{4}{n}\right)!\left(\mathrm{1103}+\mathrm{26390}{n}\right)}{\left({n}!\right)^{\mathrm{4}} \mathrm{396}^{\mathrm{4}{n}} }=\frac{\mathrm{1}}{\pi}\:\:\:\left({Prove}\:{that}\right) \\ $$
Commented by Dwaipayan Shikari last updated on 11/Oct/20
This had been done by Ramanujan. Do you have any idea  how to solve it?
$${This}\:{had}\:{been}\:{done}\:{by}\:{Ramanujan}.\:{Do}\:{you}\:{have}\:{any}\:{idea} \\ $$$${how}\:{to}\:{solve}\:{it}? \\ $$
Commented by mindispower last updated on 11/Oct/20
see generalised hyergeometric function
$${see}\:{generalised}\:{hyergeometric}\:{function} \\ $$

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