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How did Ramanujan find $\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n+41}{25920^n}=\frac{288\sqrt{5}}{5\pi}?$

The formula $$\sum_{n=0}^\infty (-1)^n\frac{(1/2)_n(1/4)_n(3/4)_n}{n!^3}\frac{644n+41}{25920^n}=\frac{288\sqrt{5}}{5\pi}$$ (in older notation) appears as eq. 38 in Ramanujan's paper Modular equations ...
Nomas2's user avatar
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2 votes
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Ellipsoidal harmonics - A Series expansion for Lame functions of the second kind

$\underline{Intro \;to \;skip}$ In the theory of ellipsoidal harmonics, Lame functions of the second kind $F_n$ arise as the second linearly independent solution (the first being Lame functions of ...
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