# 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
2 years, 2 months ago

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