$\lim_{n \rightarrow \infty} \frac{(2n)!}{n^{2n}}$ 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. 
 A: Let $x_n=\dfrac{(2n)!}{n^{2n}}$. Then
$$\begin{align*}
\frac{x_{n+1}}{x_n}&=\frac{(2\big(n+1)\big)!}{(n+1)^{2(n+1)}}\cdot\frac{n^{2n}}{(2n)!}\\\\
&=\frac{2(2n+1)}{(n+1)}\left(\frac{n}{n+1}\right)^{2n}\\\\
&=\frac{2(2n+1)}{(n+1)}\left(1-\frac1{n+1}\right)^{2(n+1)-2}\\\\
&=\frac{2(2n+1)}{(n+1)}\left(1-\frac1{n+1}\right)^{2(n+1)}\left(\frac{n+1}n\right)^2\\\\
&=\frac{2(n+1)(2n+1)}{n^2}\left(1-\frac1{n+1}\right)^{2(n+1)}\;.
\end{align*}$$
Thus,
$$\lim_{n\to\infty}\frac{x_{n+1}}{x_n}=\lim_{n\to\infty}\frac{2(n+1)(2n+1)}{n^2}\left(1-\frac1{n+1}\right)^{2(n+1)}=\frac4{e^2}<1\;,$$
and therefore $\lim\limits_{n\to\infty}x_n=0$.
A: By AM-GM we have
$(2n-2)! =2 \cdot 3 \cdot ... \cdot (2n-2) \leq  (\frac{2+3+..+2n-2}{2n-3})^{2n-3}=n^{2n-3}$
Thus
$$0 \leq \frac{(2n)!}{n^{2n}} \leq \frac{n^{2n-3}(2n-1)(2n)}{n^{2n}}=\frac{(2n-1)(2n)}{n^{3}}$$
A: Alternative (guidnece for) solution:
Consider $\sum\left( \frac{(2n)!}{n^{2n}} \right)$, that sum is converges, one can use the ratio test to show that, hence the $\left( \frac{(2n)!}{n^{2n}} \right)\to 0$  (Why?)
A: 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.
A: You definitely can't write $(2 n)!$ as $2^n n!$.
The following bound is easy to prove: $n! \le n^n$ (consider $n! = 1 \cdot 2 \cdot 3 \cdot \ldots \cdot n \le n \cdot n \cdot \ldots \cdot n = n^n$), but that isn't enough for you.
A sharper bound can be obtained by considering:
$$
(n!)^2 = (1 \cdot n) \cdot (2 \cdot (n - 1)) \cdot \ldots \cdot (n \cdot 1)
$$
We now want a bound for the factors. Take the function:
$$
f(k) = k (n + 1 - k)
$$
To get the maximum we set $f'(k) = 0$, i.e., $n + 1 - 2k = 0$, which gives $k^* = \frac{n + 1}{2}$ As $f''(k^*) = - 2 < 0$, this is the maximum. So:
$$
\begin{align*}
(n!)^2 &\le \left( \frac{n + 1}{2} \right)^{2 n} \\
n! &\le \left( \frac{n + 1}{2} \right)^n
\end{align*}
$$
This should make the rest of the proof simple.
A: 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. 
