$\lim_{n \rightarrow \infty} \frac{(2n)!}{n^{2n}}$ I am trying to show  $$\lim_{n \rightarrow \infty} \frac{(2n)!}{n^{2n}}$$ 
I tried breaking it down, and got stuck when trying to $\left( \frac{2^{n}n!}{n^{n}} \right)$ goes to 0. 

We can write
$$\frac{(2 n)!}{n^{2 n}}= \frac{(2 n)(2 n - 1) \ldots (n + 1)}{n^n}\cdot\frac{n!}{n^n}=x_n y_n\;,$$
where
$$x_n := \frac{(2 n)(2 n - 1) \ldots (n + 1)}{n^n} = \frac{n + n}{n}\cdots\frac{n + 1}{n} = \left(1 + \frac{n}{n}\right)\ldots\left(1 + \frac{1}{n}\right)$$
and
$$y_n := \frac{n!}{n^n}$$
for each $n=1, 2, 3, \ldots$. 
Now $x_n$ goes to $1$ as $n$ goes to infinity, whereas the ratio $\frac{y_{n + 1}}{y_n} = \frac{1}{(1 + \frac{1}{n})^n}$ goes to $e^{-1}<1$, which implies that $y_n$ is a monotonically decreasing sequence of positive real numbers and so converges to some non-negative quantity. 
Hence the original sequence does converge. 
