Convergence point of $\frac{\prod_{k=1}^n (2k-1)} {\prod_{k=1}^n 2k}$
HINT: $a_n < b_n \colon = \prod_{k=1}^n \frac{2k}{2k+1}$ so $a_n < \sqrt{a_n b_n} = \sqrt{\frac{1}{2n+1}}$
If you know Wallis's Product, then you can use the approximation $$\frac{\pi}{2}\approx \prod_{k=1}^n\,\left(\frac{2k}{2k-1}\cdot\frac{2k}{2k+1}\right)$$ to show that $$\frac{\pi}{2} \approx \frac{1}{2n+1}\,\prod_{k=1}^n\,\left(\frac{2k}{2k-1}\right)^2\,.$$ That is, $$\prod_{k=1}^n\,\frac{2k-1}{2k}\approx\frac{1}{\sqrt{\pi\left(n+\frac{1}{2}\right)}}\,.$$