Evaluating $\lim\left(\frac{(2n)!}{n!n^n}\right)^{1/n}$

Question

How to find: $$\lim_{n\to \infty}\left(\frac{(2n)!}{n!n^n}\right)^{\frac1n}$$

I tried as

$\displaystyle \lim\left(\frac{(2n)!}{n!n^n}\right)^{1/n}=\lim_{n \to \infty}\left(\frac{2^n(1.3.5...2n-1)}{n^n}\right)^{1/n}=\lim_{n \to \infty}\frac{2(1.3.5...2n-1)^{1/n}}{n}=2\lim_{n \to \infty}\left(\frac{1.3.5....2n-1}{n^n}\right)^{1/n}$

At the begining I used $\frac{(2n)!}{n!}=2^n(1.3.5...2n-1)$

How should I move further?

Answer 1

Call $a_n=\frac{(2n)!}{n!n^n}$. You have that, using root-ratio strategy, if they exist $$\lim \sqrt[n] {a_n} =\lim \frac {a_n} {a_{n-1}}=\lim \frac {(2n)!(n-1)!(n-1)^{n-1}}{n!n^n(2n-2)!}=\lim \frac {(2n)(2n-1)}{n^2}\left(\frac {n-1}n\right)^{n-1}=\frac 4e$$

Answer 2

$$a_n=\left(\frac{(2n)!}{n!\,n^n}\right)^{1/n}\implies \log(a_n)=\frac 1n \log\left(\frac{(2n)!}{n!\,n^n}\right)$$ $$\log\left(\frac{(2n)!}{n!\,n^n}\right)=\log((2n)!)-\log(n!)-n\log(n)$$ Using Stirling approximation and continuing with Taylor series, then $$\log\left(\frac{(2n)!}{n!\,n^n}\right)=n (2 \log (2)-1)+\frac{\log (2)}{2}-\frac{1}{24 n}+O\left(\frac{1}{n^3}\right)$$ $$\log(a_n)=(2 \log (2)-1)+\frac{\log (2)}{2n}-\frac{1}{24 n^2}+O\left(\frac{1}{n^4}\right)$$ $$a_n=e^{\log(a_n)}=\frac 4e \left(1+\frac{\log (2)}{2 n}+\frac{3 \log ^2(2)-1}{24n^2} \right)+O\left(\frac{1}{n^3}\right)$$

Answer 3

Hint:

let L be the given limit.thus it is easy to show using property of log that:

$logL=\lim_{n \to \infty}(\frac{1}{n}){\sum_{r=1}^n}{log(1+\frac{r}{n})}$

=$\int_{0}^{1}log(1+x)dx$

using integraton by parts the rest can be done easily

Answer 4

$$\lim\limits_{n \to \infty}\left(\frac{(2n)!}{n!n^n}\right)^{\frac1n} = \lim\limits_{n \to \infty}\left(\left(1+\frac{1}{n} \right)\left( 1+\frac{2}{n}\right) \cdots \left( 1+\frac{n}{n}\right)\right)^{\frac1n} = \\ =\lim\limits_{n \to \infty}e^{\frac{1}{n} \ln \left(\left(1+\frac{1}{n} \right)\left( 1+\frac{2}{n}\right) \cdots \left( 1+\frac{n}{n}\right)\right)} =\lim\limits_{n \to \infty} e^{\frac{1}{n}\sum_{i=1}^{n}\ln \left(1+\frac{i}{n} \right)} = \\ =e^{\int_{0}^{1} \ln (1+x)\,dx} = e^{2 \ln 2-1} = \frac{4}{e} $$

Answer 5

Using Stirling approximation, $$\Rightarrow \bbox[5px,border:2px solid red] {n!\approx(\frac{n}{e})^n}$$

Given,

$$\lim\limits_{n \to \infty}\left(\frac{(2n)!}{n!n^n}\right)^{\frac1n}=\lim\limits_{n \to \infty}\ \frac{(\frac{2n}{e})^{2n}} {(\frac{n}{e})^n.n}$$

$$=\lim\limits_{n \to \infty}\frac{(\frac{4n^2}{e^2})}{(\frac{n^2}{e})}$$

$$=\frac{4}{e}$$