Product of all positive odd numbers

Question

I am looking for simplest regularisation of infinite product: $$ \prod^{\infty}_{k≥0} (2k+1) = ?? $$

Answer 1

A non-rigorous methode, the answer seems correct, but because of the link in the comments I doubt everything I did, because i would make the same mistake the question asker asked there (I think).

$$\prod_{k=1}^{m}(2k-1)=\exp(\sum_{k=1}^{m}ln(2k-1))$$

$$\sum_{k=1}^{m}ln(2k-1)=\sum_{k=1}^{2m}ln(k)-\sum_{k=1}^{m}ln(2k)$$

Then based upon linearity, if regularised for m $(S\infty)$. $$\sum_{k=1}^{2m}ln(k)-\sum_{k=1}^{m}ln(k)$$ $$\sum_{k=1}^{S\infty}ln(k)-\sum_{k=1}^{S\infty}ln(k)=0 $$ And with zeta(0) $$\sum_{k=1}^{S\infty}ln(2)=\frac{-ln(2)}{2}$$

Thuse $$\sum_{k=1}^{S\infty}ln(2k-1)=\frac{ln(2)}{2}$$ $$\prod_{k=1}^{S\infty}(2k-1)=\sqrt{2}$$