Evaluating $\lim_{n \to \infty} \frac{(2n)!}{2^n (n!)^2}$ 
Can you please explain how should I evaluate this limit
$$\lim_{n \to \infty} \frac{(2n)!}{2^n (n!)^2}$$

I know the solution is $\geq1$ but I don't know how I can just simplify like this but I stuck here $$\frac{2n(2n-1)(2n-2)...1}{2^n(n(n-1)(n-2)...1)}$$ I don't know if I should simplify it like that as I said the answer is $\geq1$ but  my textbook doesn't explain why?

Taking logarithm, the sequence term becomes
$$\sum_{k=1}^{2n}\ln(k)-2\sum_{k=1}^n\ln(k)-n\ln(2)$$
$$=\sum_{k=n+1}^{2n}\ln(k)-\sum_{k=1}^n\ln(k)-n\ln(2)$$
$$=\sum_{k=1}^n\ln(\frac{k+n}{k})-n\ln(2)$$
$$n\Bigl(\frac 1n\sum_{k=1}^n\ln(1+\frac nk)-\ln(2)\Bigr)$$
The Riemann sum goes to $$\int_0^1(\ln(x+1)-\ln(x))dx=2\ln(2)+1$$
The limit will be $+\infty$.
