Compute $\lim _{n \rightarrow \infty} \frac{1}{n}\ln \frac{(2n)!}{n^n n!}$

We have
\[\lim_{n \rightarrow \infty}\frac{1}{n}\ln \frac{(2n)!}{n^n n!}=\lim _{n \rightarrow \infty} \frac{1}{n} \ln \frac{2n \times (2n-1)\times \dots \times (n+1)  \times n \times (n-1) \times \dots \times 2 \times 1}{n^n n!}\]
\[=\lim _{n \rightarrow \infty} \frac{1}{n}\ln \frac{(2n)!}{n^n n!}=\lim _{n \rightarrow \infty} \frac{1}{n} \ln \frac{2n \times (2n-1)\times \dots \times (n+1)}{n^n}\]
\[=\lim _{n \rightarrow \infty}  \frac{1}{n}\ln \left( (\frac{2n}{n}) \times (\frac{2n-1}{n})\times \dots (\frac{n+2}{n}) \times (\frac{n+1}{n}) \right)\]
\[=\lim _{n \rightarrow \infty} \frac{1}{n} \ln\left((1+\frac{1}{n})(1+\frac{2}{n}) \times \dots (1+\frac{n}{n}) \right)\]
\[=\lim _{n \rightarrow \infty} \frac{1}{n}\left( \ln(1+\frac{1}{n}) + \ln(1+\frac{2}{n}) \dots+ \ln(1+\frac{n}{n}) \right)\]
\[=\int_{1}^{2}\ln x dx.\]
Next we compute the above integral. Let $u=\ln x$ and $dv=dx$. Then $du=\frac{1}{x}$ and $v=x$. Using integration by parts formula
\[\int_{1}^{2}\ln x dx=x \ln x|_1^2-\int_1^{2}x \times \frac{1}{x}dx=2\ln 2-\ln 1-\int_1^2 dx\]
\[=2\ln 2 -1.\]
Note that $ln (1)=0.$ Hence
\[\lim_{n \rightarrow \infty}\frac{1}{n}\ln \frac{(2n)!}{n^n n!}=2\ln 2-1.\]