$\lim_{n \rightarrow \infty} \frac{(2n)!}{n^{2n}}$ 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. 

It is not true that $\left( \frac{(2n)!}{n^{2n}} \right)=\left( \frac{2^{n}n!}{n^{n}} \right)$  The left side has a factor $2n-1$ in the numerator while the right side does not.  But you can use Stirling's approximation to say $$\frac {(2n)!}{n^{2n}}\approx \frac {(2n)^{2n}}{(ne)^{2n}}\sqrt{4 \pi n}$$ and the powers of $\frac 2e$ take it to zero.
