TITLE: Solving $\lim_{n\to\infty}(\frac{(2n)!}{n^n\cdot n!})^{1/n}$
QUESTION [6 upvotes]: Find the following limit.$$\lim_{n\to\infty}\left(\dfrac{(2n)!}{n^n\cdot n!}\right)^{1/n}$$

My work:
Let assume the given limit be $y$
$$\begin{aligned}&y = \lim_{n\to\infty}\left(\dfrac{(2n)!}{n^n\cdot n!}\right)^{1/n}\\\\&\implies y= \lim_{n\to\infty}\left(\dfrac{(2n)(2n-1)(2n-2)...3.2.1}{n^n\cdot n!}\right)^{1/n}\\\\&\implies y= \lim_{n\to\infty}\left(\dfrac{\Big[(2n)(2n-2)(2n-4)...6.4.2\Big]\Big[(2n-1)(2n-3)(2n-5)..5.3.1\Big]}{n^n\cdot n!}\right)^{1/n} \\\\&\implies y= \lim_{n\to\infty}\left(\dfrac{\Big[2^n(n!)\Big]\Big[(2n-1)(2n-3)(2n-5)...5.3.1\Big]}{n^n\cdot n!}\right)^{1/n}\\\\&\implies y= \lim_{n\to\infty}\left(\dfrac{(4n-2)(4n-6)(4n-8)...(10)(6)(2)}{n^n}\right)^{1/n}\\\\&\implies \log(y)= \lim_{n\to\infty}\dfrac1n\log\left(\dfrac{(4n-2)(4n-6)(4n-8)...(10)(6)(2)}{n^n}\right)\\\\&\overset{({\Large*})}\implies \log(y)= \lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\log\left(\dfrac{4n - 2(2r-1)}{n}\right)\\\\&\overset{({\Large*})}\implies \log(y)= \lim_{n\to\infty}\dfrac1n\sum_{r=1}^n\log\left(4 - \dfrac{4r}{n} + \dfrac2n\right)\\\\&\overset{({\Large*})}\implies \log(y)= \int_0^1\log(4 - 4x ) dx\end{aligned}$$
I'm  not sure if the steps (*) are correct or not. Can anyone guide me please.

REPLY [2 votes]: 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}
$