سوال ۲۷ کنکور ارشد مهندسی کامپیوتر ۱۴۰۲

We evaluate $L=\lim_{n\to\infty}\dfrac{(n!)^{1/n}}{n}$ by taking logarithms.

$\ln L=\lim_{n\to\infty}\left[\frac{1}{n}\ln(n!)-\ln n\right]=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\big(\ln k-\ln n\big)=\lim_{n\to\infty}\frac{1}{n}\sum_{k=1}^{n}\ln\frac{k}{n}.$

The right-hand side is a Riemann sum for $\int_{0}^{1}\ln x\,dx$ (sampling at $x=k/n$ with width $1/n$):

$\int_{0}^{1}\ln x\,dx=\Big[x\ln x-x\Big]_{0}^{1}=-1.$

Hence $\ln L=-1$, so

$L=e^{-1}.$

Stirling check. Using $n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^{n}$, we get $(n!)^{1/n}\sim \frac{n}{e}\,(2\pi n)^{1/(2n)}$, and since $(2\pi n)^{1/(2n)}\to 1$,

$\frac{(n!)^{1/n}}{n}\to\frac{1}{e}=e^{-1}.$

The answer is option 2.

Exam trick: Limits of the form $\frac{1}{n}\ln(\text{product})$ usually collapse into a Riemann sum of $\ln$; recognizing $\frac{1}{n}\sum \ln(k/n)\to\int_0^1\ln x\,dx=-1$ instantly gives the $\tfrac{1}{e}$ factor that Stirling also predicts.