$1cdot3cdot5cdots(2n-1)

Question

I found that excercise in an old book without any hint nor solution. Olthough I know a thing ot two about induction, this one seems to be too tricky for me. I know for sure that it's a proof by induction
It is known that: $3e^3 < 64$ (for the base case)
Prove that:
$$1cdot3cdot5cdots(2n-1) < left( frac{2n}{e} right)^{n+1}, quad n in mathbb{N}, n ge 2$$
The induction that I am trying to do looks fine in terms of calculations, yet it is not proving anything.

Claude Leibovici · Answer

Without induction, the problem would be quite simple since you want to prove that
$$frac{2^n }{sqrt{pi }}Gamma left(n+frac{1}{2}right)<left( frac{2n}{e} right)^{n+1}$$ Taking logarithms and using Stirling approximation
$$log(text{rhs - lhs})=left(log (n)-1+frac{log (2)}{2}right)+frac{1}{24 n}+Oleft(frac{1}{n^3}right)$$
$$text{rhs - lhs}=frac{sqrt{2} n}{e}+frac{1}{12 sqrt{2} e}+frac{1}{576 sqrt{2} e
   n}+Oleft(frac{1}{n^2}right)$$

$1cdot3cdot5cdots(2n-1) < (frac{2n}{e})^{n+1}, n in mathbb{N}, n geq 2$ proof doesn't seem to work

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