Solving $\lim_{n\to\infty}(\frac{(2n)!}{n^n\cdot n!})^{1/n}$

Question

Find the following limit:$$\lim_{n\to\infty}\left(\dfrac{(2n)!}{n^n\cdot n!}\right)^{1/n}$$

My work:

Lets assume the given limit be $y$ $$\begin{align}y& = \lim_{n\to\infty}\left(\dfrac{(2n)!}{n^n\cdot n!}\right)^{1/n}\\\\&= \lim_{n\to\infty}\left(\dfrac{(2n)(2n-1)(2n-2)...2.1}{n^n\cdot n!}\right)^{1/n}\\\\&= \lim_{n\to\infty}\left(\dfrac{\Big[(2n)(2n-2)(2n-4)...4.2\Big]\Big[(2n-1)(2n-3)(2n-5)...3.1\Big]}{n^n\cdot n!}\right)^{1/n} \\\\&= \lim_{n\to\infty}\left(\dfrac{\Big[2^n(n!)\Big]\Big[(2n-1)(2n-3)(2n-5)...3.1\Big]}{n^n\cdot n!}\right)^{1/n}\end{align}$$

Taking $\log$ both sides, $$\begin{align}\log(y)&= \lim_{n\to\infty}\left(\dfrac{(4n-2)(4n-6)(4n-8)...(10)(6)(2)}{n^n}\right)^{1/n}\\\\&= \lim_{n\to\infty}\dfrac1n\log\left(\dfrac{(4n-2)(4n-6)(4n-8)...(10)(6)(2)}{n^n}\right)\\\\&\overset{({\Large*})}= \lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\log\left(\dfrac{4n - 2(2r-1)}{n}\right)\\\\&\overset{({\Large*})}= \lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\log\left(4 - \dfrac{4r}{n} + \dfrac2n\right)\\\\&\overset{({\Large*})}= \int_0^1\log(4 - 4x ) dx\end{align}$$

I'm not sure if the steps (*) are correct or not. Can anyone guide me please.

Answer 1

The easy way is by using $\dfrac{n!^{1/n}}{n} \to \dfrac1{e} $ (which is simpler than Stirling and can be proved by elementary methods).

Then

$\begin{array}\\ \left(\dfrac{(2n)!}{n^n\cdot n!}\right)^{1/n} &=\dfrac1{n}\dfrac{(2n)!)^{1/n}}{(n!)^{1/n}}\\ &=\dfrac1{n}\dfrac{((2n)!)^{1/(2n)})^2}{(n!)^{1/n}}\\ &\to\dfrac1{n}\dfrac{(2n/e)^2}{n/e}\\ &=\dfrac{4}{e}\\ \end{array} $

Answer 2

Using Stirling Approximation you get answer $\boxed{\frac{4}{e}}$

P.S:

$n! \underset{n \to \infty}{\sim} \sqrt{2 \pi n} \cdot \left( \frac{n}{e} \right)^n$

$(2n)! \underset{n \to \infty}{\sim} \sqrt{4 \pi n} \cdot \left( \frac{2n}{e} \right)^{2n}$

Answer 3

It is a well known result in Calculus that $$\liminf_n\frac{|a_{n+1}|}{|a_n|}\leq\liminf_n\sqrt[n]{|a_n|}\leq\limsup_n\sqrt[n]{|a_n|}\leq\limsup_n\frac{|a_{n+1}|}{|a_n|}$$ With $a_n=\frac{(2n)!}{n^n\, n!}$ one gets $$\frac{a_{n+1}}{a_n}=\frac{(2n+2)(2n+1)}{(n+1)^2}\big(\frac{n}{n+1}\big)^n$$ Can you finish from here?