Prove $\lim_{n\to\infty}\frac{2\cdot4\cdot6\cdot...\cdot(2n-2)(2n)}{1\cdot3\cdot5\cdot...\cdot(2n-1)}\frac{1}{\sqrt{2n+1}}=\sqrt\frac{\pi}{2}$

Using the double factorial notation we need to find $$\lim_{n\to\infty} \frac {(2n)!!}{(2n-1)!!\sqrt {2n+1}}$$

Now using the relation between double factorial and the factorial, the limit changes to $$\lim_{n\to\infty} \frac {2^{2n}(n!)^2}{(2n)!\sqrt {2n+1}}$$

Using Stirling's approximation for factorials we get $$\lim_{n\to\infty} \frac {2^{2n}\cdot (2\pi n)\cdot \left(\frac ne \right)^{2n}}{\sqrt {2\pi}\cdot\sqrt {2n} \cdot\left(\frac {2n}{e}\right)^{2n} \cdot \sqrt {2n+1}}$$

Hence limit changes to $$\lim_{n\to\infty} \frac {n\sqrt {2\pi}}{\sqrt {2n} \cdot \sqrt {2n+1}}$$

Which easily evaluates to $\sqrt {\frac {\pi}{2}}$

Have a look here: Wallis' integrals and here.