\limsup_{n\to\infty}\frac{\frac{(2n)!}{(n!)^2}}{\frac{4^n}{\sqrt n}}=\limsup_{n\to\infty}\frac{\left(\frac{2n}e\right)^{2n}\sqrt{4\pi n}\cdot\sqrt n}{4^n\cdot\left(\frac n e\right)^{2n}2\pi n}=\limsup_{n\to\infty}\frac{\left(\frac{2n}e\right)^{2n}}{\left(\frac{2n}e\right)^{2n}\sqrt{\pi}}=\frac 1{\sqrt{\pi}}