Proof that $\frac{2\times2\times2\times4\times4\cdots(2n)\times(2n)}{3\times3\times5\times5\cdots(2n-1)\times(2n+1)}$ converges to $\frac{\pi}{2}$

Question

I defined $$S_{2n}=\frac{1\cdot 3\cdot 5\cdots (2n-3)\cdots (2n-1) }{2\cdot 4\cdot 6\cdots (2n-2)\cdot (2n)}\cdot \frac{\pi}{2}$$ and $$S_{2n+1}=\frac{2\cdot 4\cdot 6\cdots (2n-2)\cdots (2n) }{1\cdot 3\cdot 5\cdots (2n-1)\cdot (2n+1)},$$ where $$S_n=\int_{0}^{\frac{\pi}{2}} (\sin x)^n\, dx.$$ I proved that $S_n=\left( \frac{n-1}{n}\right) S_{n-2}$ and that $S_{n}\geq S_{n+1}$ for all $n\in \mathbb{N}$. Now, I need that $$W_n=\frac{2\times2\times2\times4\times4\cdots(2n)\times(2n)}{3\times3\times5\times5\cdots(2n-1)\times(2n+1)}=\frac{S_{2n+1}}{S_{2n}}\cdot\frac{\pi}{2}$$ converges to $\frac{\pi}{2}$. I tried to prove that $$\frac{S_{2n+1}}{S_{2n}}=\frac{2\times2\times2\times4\times4\cdots(2n)\times(2n)}{3\times3\times5\times5\cdots(2n-1)\times(2n+1)}\cdot\frac{2}{\pi}\xrightarrow{n\to\infty}1,$$ but I did not succeed.