Proof: $\lim\limits_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=e^{-1}$

Question

Prove that $\displaystyle\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=e^{-1}$.

Here's my solution:

Stirling's formula tells us $$\lim_{n\to\infty}\frac{n!}{n^ne^{-n}\sqrt{2\pi n}}=1$$which implies $$\lim_{n\to\infty}\sqrt[n]{\frac{n!}{n^ne^{-n}\sqrt{2\pi n}}}=1$$then simplifying the left side we have $$\lim_{n\to\infty}\sqrt[n]{\frac{n!}{n^n}}\lim_{n\to\infty}\sqrt[n]{e^{n}}\lim_{n\to\infty}\frac{1}{\sqrt[2n]{2\pi n}}=e\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=1$$ since $\lim_{n\to\infty}\sqrt[2n]{2\pi n}=1$. Divide both sides by $e$ and we're done.

Is this correct? This is a problem from Problems in Real Analysis by Radulescu and Andreescu. The book gives two other proofs. Thanks!

Answer 1

You could make it shorter $$A=\frac{\sqrt[n]{n!}}n\implies \log(A)=\frac 1n \log(n!)-\log(n)$$ Using Stirling approximation$$ \log(n!)=n (\log (n)-1)+\frac{1}{2} \left(\log (2 \pi )+\log \left(n\right)\right)+\frac{1}{12 n}+O\left(\frac{1}{n^2}\right)$$ leads to $$\log(A)=-1+\frac{\log (n)+\log (2 \pi )}{2 n}+\frac{1}{12 n^2}+O\left(\frac{1}{n^3}\right)$$

Answer 2

Let $$A_n=\frac{\sqrt[n]{n!}}n.$$ Then $$\log(A_n)=\frac 1n \log(n!)-\log(n)=\frac 1n \sum_{k=1}^n\log k-\log(n)=\sum_{k=1}^n\frac1n\log(\frac{k}{n})\to\int_0^1\log xdx=-1$$ and hence $$ \lim_{n\to\infty} A_n=e^{-1}. $$