Mathematics - Definite Integration Question with Solution | TestHub

Let $P=\frac{(n!{)}^{\frac{1}{n}}}{n}$.

Then $\ln P=\frac{1}{n}\ln (n!)-\ln n=\frac{1}{n}{\sum }_{r=1}^n\ln r-\ln n=\frac{1}{n}{\sum }_{r=1}^n\ln (\frac{r}{n})$.

As $n\to \infty$, this becomes a Riemann sum: ${\int }_0^1\ln xdx$. Integrating by parts: $[x\ln x-x{]}_0^1=(1\ln 1-1)-{\lim }_{x\to 0}(x\ln x-x)=-1-0=-1$.

Therefore, ${\lim }_{n\to \infty }\ln P=-1$, so ${\lim }_{n\to \infty }P=e^{-1}$.

Final Answer: The final answer is $\boxed{e^{-1}}$