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Dec 30, 2020 at 11:45 comment added Roman Thanks to Dr. Wolfgang Hintze and J.M. for putting this calculation in its proper mathematical frame.
Dec 30, 2020 at 9:28 comment added J. M.'s missing motivation @Dr. and Roman, indeed, using Gauß's hypergeometric theorem is a clever idea...
Dec 30, 2020 at 8:48 history edited Roman CC BY-SA 4.0
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Dec 29, 2020 at 21:57 comment added Dr. Wolfgang Hintze @ Roman Very good idea! The general formula is $_2F_1(a,b,c,1) = \frac{\Gamma(c-b-a) \Gamma(c)}{\Gamma(c-a)\Gamma(c-b)}$. Unfortunately, my Mathematica was not able to verify this formula.
Dec 23, 2020 at 20:01 history edited Roman CC BY-SA 4.0
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Dec 23, 2020 at 18:06 history answered Roman CC BY-SA 4.0