Square your limit problem

Use Stirling's Approximation to re-express this as follows

$\sqrt [ n ]{ \dfrac { \left( 2n \right) ! }{ n!{ n }^{ n } } } =\sqrt [ n ]{ \dfrac { \sqrt { 2\pi 2n } { \left( \dfrac { 2n }{ e } \right) }^{ 2n } }{ \sqrt { 2\pi n } { \left( \dfrac { n }{ e } \right) }^{ n } n ^{ n } } } ={ \left( \sqrt { 2 } \right) }^{ \frac { 1 }{ n } }\left( \dfrac { 4 }{ e } \right)$

As $n\rightarrow \infty$ , this becomes

${ \left( \sqrt { 2 } \right) }^{ \frac { 1 }{ n } }\left( \dfrac { 4 }{ e } \right) =\left( \dfrac { 4 }{ e } \right) =1.471568764...$

The logarithm of the given expression $a_n=\sqrt[n]{...}$ is $\ln(a_n)=\sum_{k=1}^{n}\ln\left(1+\frac{k}{n}\right)\frac{1}{n}$ , a Riemann sum of $\ln(x)$ from 1 to 2. Thus $\lim_{n\to\infty}\ln(a_n)=\int_{1}^{2}\ln(x)dx=2\ln(2)-1$ and $\lim_{n\to\infty}a_n=e^{2\ln(2)-1}=\boxed{\frac{4}{e}}$

I'm using approximation for central binomial coefficient $\dfrac{(2n)!}{n! n^n} = \dfrac{(2n)!n!}{(n!)^2 n^n} =\dfrac{4^n}{\sqrt {2\pi n}}\cdot \dfrac{ {\sqrt{ 2\pi n }n^n}}{e^n}=\dfrac{4^n \cdot n^n}{e^n}$ Gives us limit $\lim_{n\to \infty} \left(\dfrac{(2n)!}{n! n^n}\right)^{\frac{1}{n}}= \dfrac{4}{e}$