Prove $\lim_{n\to\infty}\frac{2\cdot4\cdot6\cdot...\cdot(2n-2)(2n)}{1\cdot3\cdot5\cdot...\cdot(2n-1)}\frac{1}{\sqrt{2n+1}}=\sqrt\frac{\pi}{2}$ I have to prove that the following limit is equal to $\sqrt{\pi/2}$:
$$\lim_{n\to\infty}\frac{2\cdot4\cdot6\cdot...\cdot(2n-2)(2n)}{1\cdot3\cdot5\cdot...\cdot(2n-1)}\frac{1}{\sqrt{2n+1}}=\sqrt\frac{\pi}{2}$$
In order to calculate this limit, we know that:
$$I_n=\int_0^{\frac{\pi}{2}}\sin^nx\ dx\quad I_{2n}=\frac{1\cdot3\cdot..\cdot(2n-3)(2n-1)}{2\cdot4\cdot..\cdot(2n-2)(2n)}\frac{\pi}{2}\quad I_{2n+1}=\frac{2\cdot4\cdot..\cdot(2n-2)(2n)}{1\cdot3\cdot..\cdot(2n-1)(2n+1)}$$
I have tried to rewrite the limit as:
$$\lim_{n\to\infty}\frac{1}{I_{2n}\sqrt{2n+1}}\frac{\pi}{2}$$
But I don't know how to continue... Could you help me? Thanks in advance!

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 $\sqrt {\frac {\pi}{2}}$
