Evaluate $\prod \frac {2k - 1} {2k}$ [duplicate]

$$\prod_{k=1}^{n}\frac{2k-1}{2k}=\frac{2-1}{2}\cdot\frac{4-1}{4}\cdot\frac{6-1}{6}\cdot\dots\cdot\frac{2n-1}{2n}=$$ $$\frac{1}{2}\cdot\frac{3}{4}\cdot\frac{5}{6}\cdot\dots\cdot\frac{2n-1}{2n}=\frac{1\cdot3\cdot5\dots\cdot(2n-1)}{2\cdot4\cdot6\dots\cdot2n}=\frac{\frac{2^n\Gamma\left(n+\frac{1}{2}\right)}{\sqrt{\pi}}}{2^n\Gamma\left(n+1\right)}=\frac{\Gamma\left(n+\frac{1}{2}\right)}{n!\sqrt{\pi}}$$

I tried a different and definitely a far more elementary approach and, surprisingly, it gave a better estimate. We have obviously (which I had overlooked!)

$$\prod_{k = 1}^{n} \frac {2k - 1} {2k} = \frac {(2n)!} {4^n (n!)^2}$$

which, by Stirling formula, is

$$=\frac {1} {\sqrt {\pi n}} \exp \left (\frac {1} {192 n^3} - \frac {1} {8n}\right).$$

This estimate is indeed very precise, since for the very small value $n = 5$ the actual value is $0.24609375$ and the estimated value is $0.2460938695063\cdots$.