Help with Limit: $\lim_{n \to \infty} \frac{(2 n)! (n)^n}{n! (2 n)^{2 n}}$

Question

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.

Answer 1

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, $$\displaystyle\lim_{n\to\infty} \dfrac{(2n)!n^n}{n!(2n)^{2n}}=0$$

Answer 2

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.

Answer 3

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ \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}

Answer 4

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!

Answer 5

\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).

Answer 6

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