Help with Limit: $\lim_{n \to \infty} \frac{(2 n)! (n)^n}{n! (2 n)^{2 n}}$ Honestly, I am just plain stuck, I have been hitting my head against it for $2$ days straight. I know the solution should be $2^{-2}$ but...
Help would be appreciated. If anyone has a Wolfram Alpha PRO account I am sure that thing churns out the solution as it is something quite basic I just don't see it and its driving me crazy.
Edit: Stacks seemed to correct my formating from this sloppy one "Limit[(2 n)! n^n (E^n/(n! (2 n)^(2 n))), n -> Infinity]" $\displaystyle\lim_{n \to \infty} \frac{(2 n)! e^n (n)^n}{n! (2 n)^{2 n}}$
to this one $\displaystyle\lim_{n \to \infty} \frac{(2 n)! (n)^n}{n! (2 n)^{2 n}}$ somewhere dropping the $e^n$ term. I should have learned to propperly use latex in stackexchange first.
 A: Let $$(2n-1)!!=1\cdot3\cdot 5\cdot\ldots\cdot(2n-1) $$
$$\dfrac{(2n)!n^n}{n!(2n)^{2n}}=\dfrac{(2n-1)!!\cdot(2^nn!)\cdot n^n}{2^{2n} n^{2n} n!}=\dfrac{(2n-1)!!}{2^{n} n^{n}}\le\dfrac{2^n\cdot n!}{2^n n^n}=\dfrac{n!}{n^n}$$Now since $\displaystyle\lim_{n\to\infty} \dfrac{n!}{n^n}=0$ we have, $$\displaystyle\lim_{n\to\infty} \dfrac{(2n)!n^n}{n!(2n)^{2n}}=0$$
A: When I see a problem of limits with factorials, I immediately think about Stirling approximation
$$\log(p!)=p (\log (p)-1)+\frac{1}{2} \left(\log (2 \pi )+\log
   \left(p\right)\right)+O\left(\frac{1}{p}\right)$$
$$A_n=\frac{(2 n)! (n)^n}{n! (2 n)^{2 n}}\implies \log(A_n)\sim\log((2n)!)+n\log(n)-\log(n!)-2n\log(2n)$$ Applying the formula and simplifying leads to $$\log(A_n)\sim \frac{\log (2)}{2}-n$$ from which you can easily conclude.
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,}
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\begin{align}
\lim_{n \to \infty}{\pars{2n}!\, n^{n} \over n!\,\pars{2 n}^{2n}} & =
\lim_{n \to \infty}{\bracks{\root{2\pi}\pars{2n}^{2n + 1/2}\expo{-2n}}n^{n} \over \bracks{\root{2\pi}n^{n + 1/2}\expo{-n}}\pars{2 n}^{2n}}\quad
\pars{~\begin{array}{l}
\mbox{by using}
\\
Stirling\ Asymptotic\ Expansion\end{array}~}
\\[5mm] & =
\lim_{n \to \infty}{\pars{2n}^{1/2}\expo{-n} \over n^{1/2}} =
\root{2}\lim_{n \to \infty}\expo{-n} = \bbx{\ds{0}}
\end{align}
A: Let $a_n=\frac{(2n)!\,n^n}{n!\,(2n)^{2n}}$.  Then, we can write
$$\begin{align}
\log(a_n)&=\log((2n)!)+n\log(n)-\log(n!)-2n\log(2n)\\\\
&=\sum_{k=n+1}^{2n}\log(k/n)-2n\log(2)\\\\
&= \log(2)+\sum_{k=n}^{2n-1}\log(k/n)-2n\log(2)\\\\
&\le \log(2)+\int_n^{2n}\log(x/n)\,dx-2n\log(2)\\\\
&=\log(2)-n\\\\
\end{align}$$
Since $\lim_{n\to \infty}\log(a_n)=-\infty$, then the limit of interest is
$$\lim_{n\to \infty}a_n=\lim_{n\to \infty}e^{\log(a_n)}=0$$
And we are done!
A: \begin{align*}
\lim_{n \to \infty} \frac{ (2n)! n^n }{n! (2n)^{2n}} &= \lim_{n \to \infty} \frac{ 2n\cdot(2n-1) \cdots (n+1) \cdot n! \cdot  n^n }{n! 2^{2n} n^{2n}}\\
 &= \lim_{n \to \infty} \frac{ 2n\cdot(2n-1) \cdots (n+1)  }{ 2^{2n} n^{n}}
\end{align*}
This is the limit after a little bit of simplification. What can we say about the product on the top? 
Edit
At least if I'm reading your problem correctly (there have been a few differing edits made to it) the limit should actually end up being $0$ (per Mathematica). 
A: As @erfink points out, the limit can be trivially rearranged to get
$$\lim_{n \to \infty} \frac{ 2n\cdot(2n-1) \cdots (n+1)  }{ 2^{2n} n^{n}}$$
$$=\lim_{n \to \infty} \frac{ 2n\cdot(2n-1) \cdots (n+1)  }{(4n)^n}$$ 
From here I will show an intuitive understanding of how to complete the argument.
$$2n\cdot(2n-1) \cdots (n+1)$$
$$=(n+1)(n+2)\cdots(n+n)$$ 
And so we see that there are $n$ terms in the numerator, as in the denominator. Thus, if we convert the limit to to a product of limits we get
$$=\lim_{n \to \infty}\frac{n+1}{4n}\lim_{n \to \infty}\frac{n+2}{4n}\cdots\lim_{n \to \infty}\frac{n+n}{4n}$$
Clearly all limits will be finite, and in fact all of the multiplied terms will all be less than $1$. Since there are infinitely many this must tend to zero (I leave this proof to the OP to formalize!)  
Note: @dvix notes in the comments that the above can be seen a bit more rigorously if we note that each factor $\frac{n+k}{4n}\leq\frac 12$ for all $1\leq k \leq n$, so the product is $\leq \frac{1}{2^n}$. This is a tighter argument as it gives a bound for a given $n$ as opposed to my simpler "infinite terms less than $1$ (and greater than zero) equal $0$". Clearly $\frac{1}{2^n}$ goes to zero as $n$ goes to infinity 
