Can someone please help me with this I know that we should use the ratio test but can you show me the steps in detail $$\sum_{n=1}^{\infty} \frac{(2^n n!)^2}{(2n)^{2n}}$$
And I stuck here $$\lim_{n \to \infty}\left|\frac{(2^{n+1}(n+1)!)^2}{(2n+2)^{2n+2}}.\frac{(2n)^{2n}}{(2^n n!)^2}\right|$$ $$=\lim_{n \to \infty} \left|\frac{(2n)^{2n}}{(2n+2)^{2n}}\right|$$ how should I proceed from here?
In order to use the ratio test, we have to evaluate the limit: $$ L := \lim_{n \rightarrow \infty} \left | \frac{a_{n+1}}{a_n} \right | $$ If $L <1$, the series converges; if $L >1$, it diverges; if $L=1$ or the limit does not exist, we cannot conclude anything with the ratio test. Now consider the first series; we have: $$ \lim_{n \rightarrow \infty} \left | \frac{a_{n+1}}{a_n} \right | = \lim_{n \rightarrow \infty} \left | \frac{(2^{n+1} (n+1)!)^2 (2n)^{2n} }{(2^n n!)^2 (2n+2)^{2n+2}} \right | = \frac{1}{e^2} < 1$$ Thus, the first series converges. Similarly, you can see that the second series diverges.
EDIT: to evaluate the limit, rewrite it as follows: $$ \lim_{n \rightarrow \infty} \frac{2^{2n+2}(n+1)! (n+1)! 2^{2n} n^{2n}}{2^{2n} n! n! 2^{2n+2} (n+1)^{2n+2}} = \lim_{n \rightarrow \infty} \left ( \frac{n}{n+1} \right )^{2n} =\lim_{n \rightarrow \infty} \left (1- \frac{1}{n+1} \right)^{2n} = \frac{1}{e^2} $$ where at the first passage we have simplified, and in the last one we have used a well known limit for the exponential.