Nice exercise to review Riemann sum

Exercise:

Evaluate the following $\underset{n\to \infty }{\mathop{\lim }}\,\frac{\sqrt[{{n}^{2}}]{\prod\nolimits_{k=1,2,...,n}{k!}}}{\sqrt{n}}$ 

Solution: Let $f\left( n \right)=\frac{\sqrt[{{n}^{2}}]{\prod\nolimits_{k=1,2,..,n}{k!}}}{\sqrt{n}}=\frac{{{\left( \prod\nolimits_{k=1,2,..,n}{k!} \right)}^{1/{{n}^{2}}}}}{{{n}^{1/2}}}$ 

So that, $\ln f\left( n \right)=\frac{1}{{{n}^{2}}}\ln \left( \prod\nolimits_{k=1,2,..,n}{k!} \right)-\frac{1}{2}\ln n$
Notice that, $g\left( n \right)=\ln \left( \prod\nolimits_{k=1,2,..,n}{k!} \right)=\sum\nolimits_{k=1,2,..,n}{\ln k!}=\ln 1!+\ln 2!+...+\ln n!$
Moreover, we know that, $\ln n!=\ln \left( n\left( n-1 \right)\times ....2\times 1 \right)=\ln n+\ln \left( n-1 \right)+...+\ln 2+\ln 1=\sum\nolimits_{k=1}^{n}{\ln k}$
So that, $g\left( n \right)=\underbrace{\ln 1}_{\ln 1!}+\underbrace{\ln 1+\ln 2}_{\ln 2!}+\underbrace{\ln 3+\ln 2+\ln 1}_{\ln 3!}+....+\underbrace{\sum\nolimits_{k=1}^{n}{\ln k}}_{\ln n!}$ 
                  $=n\ln 1+\left( n-1 \right)\ln 2+\left( n-2 \right)\ln 3+....+\ln n$
Hence $g\left( n \right)=n\ln 1-n\ln n+n\ln n+\left( n-1 \right)\ln 2-\left( n-1 \right)\ln n+\left( n-1 \right)\ln n+...+\ln n-\ln n+\ln n$       
   $=n\left( \ln 1-\ln n \right)+\left( n-1 \right)\left( \ln 2-\ln n \right)+....+\ln \left( n/n \right)+\frac{n\left( n+1 \right)}{2}\ln n$
Thus, $\frac{1}{{{n}^{2}}}g\left( n \right)=\frac{1}{n}\ln \left( 1/n \right)+\frac{n-1}{{{n}^{2}}}\ln \left( 2/n \right)+....+\frac{1}{{{n}^{2}}}\ln \left( n/n \right)+\frac{n+1}{2n}\ln n$
                    $=\frac{1}{n}\left( \ln \frac{1}{n}+\frac{n-1}{n}\ln \frac{2}{n}+.....+\frac{1}{n}\ln \frac{n}{n} \right)+\frac{n+1}{2n}\ln n$
So that, $\ln f\left( n \right)=\frac{1}{n}\sum\nolimits_{k=1,..,n}{\frac{n-\left( k-1 \right)}{n}\ln \frac{k}{n}}+\frac{n+1}{2n}\ln n-\frac{1}{2}\ln n$
                       $=\frac{1}{n}\sum\nolimits_{k=1,..,n}{\frac{n-\left( k-1 \right)}{n}\ln \frac{k}{n}+\frac{1}{2n}\ln n}$ 
Hence $\underset{n\to \infty }{\mathop{\lim }}\,\ln f\left( n \right)={{I}_{n}}+\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{2n}\ln n$ where ${{I}_{n}}=\underset{n\to \infty }{\mathop{\lim }}\,\frac{1}{n}\sum\nolimits_{k=1}^{n}{\frac{n-\left( k-1 \right)}{n}\ln \frac{k}{n}}$ 
Notice that, this sum is Riemann sum with $\Delta x=\frac{b-a}{n}=\frac{1-0}{n}$ and ${{x}_{i}}=a+i\Delta x$
Hence our partition will be like $0={{x}_{0}}<{{x}_{1}}<{{x}_{2}}<....<{{x}_{n}}=1$
Moreover, $\frac{n-k+1}{n}\ln \left( k/n \right)=\left( 1-\frac{k}{n}+\frac{1}{n} \right)\ln \left( k/n \right)=\left( 1-{{x}_{i}} \right)f\left( {{x}_{i}} \right)\Delta {{x}_{i}}$
Therefore, ${{I}_{n}}=\int_{0}^{1}{\left( 1-x \right)\ln x\,dx}=-\frac{3}{4}$ and $\underset{n\to \infty }{\mathop{\lim }}\,\frac{\ln n}{2n}=0$ 
Therefore, $\underset{n\to \infty }{\mathop{\lim }}\,f\left( n \right)={{e}^{-3/4}}$