Closed form for ∏∞n=12n√Γ(2n+12)Γ(2n)\prod_{n=1}^\infty\sqrt[2^n]{\frac{\Gamma(2^n+\frac{1}{2})}{\Gamma(2^n)}}

Is there a closed form for the following infinite product?
n=12nΓ(2n+12)Γ(2n)

Answer

The beautiful idea of Raymond Manzoni can actually be made rigorous. Consider a finite product Ln=1 and take its logarithm. After using duplication formula for the gamma function and telescoping, it simplifies to the following:
Ln=112nlnΓ(2n+12)Γ(2n)=(12L)ln(2π)2Lln2+212L+1lnΓ(2L+1).
This is an exact relation, valid for any L. Now it suffices to use Stirling,
1NlnΓ(N)=lnN1+O(lnNN)asN
to get
n=112nlnΓ(2n+12)Γ(2n)=ln(2π)+2(ln21)=ln8πe2.
So the answer is indeed 8πe2.

Attribution
Source : Link , Question Author : Laila Podlesny , Answer Author : Start wearing purple

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