Determine whether $\sum_n(2n!)/(n!(2n)^n)$ converges or diverges The question asks to determine whether the following series converges or diverges: 
$$\sum_{n=1}^\infty \frac{(2n)!}{n!(2n)^n}$$
applying the ratio test:
$$\begin{align} \lim_{n \to \infty} | \frac{a_{n+1}}{a_n}| &= \lim_{n \to \infty} |\frac{(2(n+1))!}{(n+1)!(2(n+1))^{n+1}} \frac{n!(2n)^n}{(2n)!}| \\ &= \frac{1}{2}\lim_{n \to \infty} |\frac{2(n+1)n!}{(n+1)n!2^n(n+1)^{n+1}} \frac{n!2^nn^n}{2(n+1)n!}| \\ &= \frac{1}{2} \lim_{n \to \infty} | \frac{n^n}{(n+1)^{n+2}} | \end{align}$$
from here i am not sure what to do. Following the limit is a term of the form $\frac{\infty^\infty}{\infty^\infty}$ and i am finding this problematic. 
Thanks in advance, 
edit: 
as cronos2 kindly points out $(2(n+1)!) \text{ is not } 2(n+1)n! \text{ but rather } (2n+2)(2n+1)(2n)!$ so making this correction we then have
$$\begin{align} &= \lim_{n \to \infty} |\frac{(2n+2)(2n+1)(2n)!}{(n+1)(n)!(2n+1)^{n+1}} \frac{(n)!(2n)^n}{(2n)!}| \\ &= \lim_{n \to \infty} \frac{(2n+1)(2n)^n}{(2n+2)^n} \\ &=\lim_{n \to \infty} |(2n+1)(\frac{2n}{2n+2})^n| \\ &=\lim_{n \to \infty} |(2n+1)(\frac{1}{1+\frac{1}{n}})^n| \\ &= \lim_{n \to \infty} |\frac{(2n+1)}{(1+\frac{1}{n})^n}| \\ &= \infty ? \end{align}$$
My answer book states that this series is convergent and not divergent? 
 A: With Stirling's estimate, $$\frac{(2n)!}{n!(2n)^n}\sim\sqrt 2\left(\frac{4}{2e} \right)^n$$
Since $\frac{4}{2e}<1$, the series converges by the limit comparison test.
A: By elementary means:
For $n=4$, after simplification,
$$\frac{\color{green}5\cdot\color{green}6\cdot7\cdot8}{\color{green}8\cdot\color{green}8\cdot8\cdot8}<\frac{6\cdot6}{8\cdot8}.$$
More generally, among the $n$ factors, those in the first half do not exceed $\dfrac34$ and the remaining ones do not exceed $1$.
A: Using the ratio test, we have
\begin{align}\frac{a_{n+1}}{a_n}&=\frac{(2n)!(2n+2)(2n+1)}{n!(n+1)(2n+2)^{n+1}}\cdot\frac{n!(2n)^n}{(2n)!}\\
&=\cdots=\frac{(2n+1)n^n}{(n+1)^{n+1}}.\end{align}
Taking the limit yields:
\begin{align}
\underset{n\to\infty}\lim\frac{(2n+1)n^n}{(n+1)^{n+1}}&=\underset{n\to\infty}\lim\frac{(2n)n^n}{(n+1)^{n+1}}\\
&=2\underset{n\to\infty}\lim\frac{n^{n+1}}{(n+1)^{n+1}}\\
&=2\underset{n\to\infty}\lim\left(\frac1{1+1/n}\right)^{n+1}\\
&=2\underset{n\to\infty}\lim\left(1+\frac 1{n-1}\right)^{-n}=\frac 2e<1.
\end{align}
