Nice Exercise mixed two ideas in one exercise Asked in the Brilliant.org

Exercise:

Show that $\underset{n\to \infty }{\mathop{\lim }}\,{{\left( \frac{n!}{{{n}^{n}}} \right)}^{1/n}}=\frac{1}{e}$

Solution: Let $y\left( n \right)={{\left( \frac{n!}{{{n}^{n}}} \right)}^{1/n}}$ Take Logarithm both sides to get

$\ln y\left( n \right)=\frac{1}{n}\ln \left( \frac{n!}{{{n}^{n}}} \right)\Rightarrow y\left( n \right)={{e}^{\frac{1}{n}\ln \left( \frac{n!}{{{n}^{n}}} \right)}}={{e}^{f\left( n \right)}}$Where $f\left( n \right)=\frac{1}{n}\ln \left( \frac{n!}{{{n}^{n}}} \right)$  

But $\frac{n!}{{{n}^{n}}}=\frac{1\times 2\times .....\times \left( n-1 \right)\times n}{n\times n\times ....\times n}$

So $\ln \left( \frac{n!}{{{n}^{n}}} \right)=\ln \left( \frac{n\left( n-1 \right)\times .....\times 2\times 1}{n\times n\times .....\times n} \right)=\ln \left( \frac{n}{n} \right)+\ln \left( \frac{n-1}{n} \right)+....+\ln \left( \frac{2}{n} \right)+\ln \left( \frac{1}{n} \right)$

$\Rightarrow \ln \left( \frac{n!}{{{n}^{n}}} \right)=\sum\limits_{i=1}^{n}{\ln \left( \frac{i}{n} \right)}$  thus $f\left( n \right)=\frac{1}{n}\ln \left( \frac{n!}{{{n}^{n}}} \right)=\frac{1}{n}\sum\limits_{k=1}^{n}{\ln \left( \frac{k}{n} \right)}$

But ${{x}_{k}}^{*}=\frac{k}{n}\,\,\And \,\,\Delta {{x}_{k}}^{*}=\frac{1-0}{n}$

So $f\left( n \right)=\Delta {{x}_{k}}^{*}\sum\limits_{k=1}^{n}{\ln {{x}_{k}}^{*}=\int_{0}^{1}{\ln x\,dx}}=\left[ x\ln x-x \right]_{0}^{1}=\ln 1-1=-1$

Hence $\underset{n\to \infty }{\mathop{\lim }}\,y\left( n \right)={{e}^{\underset{n\to \infty }{\mathop{\lim }}\,f\left( n \right)}}={{e}^{-1}}=\frac{1}{e}$  (since Exponential is continuous function)