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Jan 1, 2019 at 20:03 comment added Ester Mariucci That's a very quick way to prove it, thanks! This proof is of particular interest to me because I was also aiming to prove \Gamma(\alpha) \geq \alpha^{\alpha+1}, which indeed is essentially equivalent to the inequality on Beta by using Stirling on both \Gamma(x\alpha) and \Gamma((x+1)\alpha).
Jan 1, 2019 at 19:54 vote accept Ester Mariucci
Jan 1, 2019 at 19:54
Jan 1, 2019 at 19:45 history answered esg CC BY-SA 4.0