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Solution
\[\begin{aligned}
\left(\frac{n !}{n^{n}}\right)^{\frac{1}{n}}&=\left(\prod_{k=1}^{n} \frac{k}{n}\right)^{\frac{1}{n}}\\\\
&=e^{\frac{1}{n} \ln \left(\prod_{k=1}^{n} \frac{k}{n}\right)}\\\\
&=e^{\frac{1}{n} \sum_{k=1}^{n} \ln \left(\frac{k}{n}\right)}
\end{aligned}\]
Using Riemann summation method, we get
\[\begin{aligned}
\lim _{n \rightarrow+\infty} \frac{1}{n} \sum_{k=1}^{n} \ln \left(\frac{k}{n}\right)&=\int_{0}^{1} \ln (x) d x\\\\
&=[x \ln (x)-x]_{0}^{1}=-1
\end{aligned}\]
\[\begin{aligned}
\Rightarrow \lim _{n \rightarrow+\infty}\left(\frac{n !}{n^{n}}\right)^{\frac{1}{n}}=e^{-1}=\frac{1}{e}
\end{aligned}\]
Home -> Solved problems -> Calculate the limit
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Home -> Solved problems -> Calculate the limit