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#ramanujan

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@math_stackexchange How did Ramanujan find $\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n+41}{25920^n}=\frac{288\sqrt{5}}{5\pi}?$ #Mathematics #Ramanujan #Series
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Apr 13, 2024
@math_stackexchange Integral In Ramanujan's Letter To G.H. Hardy #Mathematics #HistoryOfMath #Ramanujan
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Dec 4, 2022