Evaluating $\lim_{n \to \infty} \frac{(2n)!}{2^n (n!)^2}$

Question

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?

Answer 1

Let $a_n=\frac{(2n)!}{2^n (n!)^2}$ and consider the limit of the ratio $$\frac{a_{n+1}}{a_n}=\frac{(2n+2)!}{2^{n+1} ((n+1)!)^2}\cdot \frac{2^n (n!)^2}{(2n)!}=\frac{(2n+2)(2n+1)}{2 (n+1)^2}\to \ ?$$ Do you know any theorem about limit of sequences which involves such ratio?

Answer 2

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 $+\infty$.