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Hard product!

\[ \large \displaystyle\prod _{ n=1 }^{ \infty }{ \dfrac { 1 }{ e } \left( \dfrac { 2n }{ 2n-1 } \right) ^{ (4n-1)/2 } } =\dfrac { { A }^{ 3 } }{ { \pi }^{ 1/4 }{ 2 }^{ 7/12 } } \]

Prove that the equation above holds true, where \(A\) denotes the Glaisher–Kinkelin constant.


This is a part of the set Formidable Series and Integrals

Note by Hummus A
5 months, 2 weeks ago

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