5

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!

Martin Sleziak
  • 53,687
Shaun_the_Post
  • 500
  • 2
  • 8
  • The proof is correct, but it feels like one can show this even without Stirling – Hagen von Eitzen Jan 22 '17 at 21:16
  • 1
    Related: http://math.stackexchange.com/q/201906/5531 – Antonio Vargas Jan 23 '17 at 15:30
  • Technically, there is a small issue here. Before breaking the limit in 3 factors, you have to explain what you can do so (it is easy, but needs to be done). Writing $\lim_n(a_nb_n)=\lim_n a_n\cdot \lim_n b_n$ presupposes that all limits have been shown to exist. – Clement C. Jan 23 '17 at 20:57

2 Answers2

5

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 $$ \lim_{n\to\infty} A_n=e^{-1}. $$

xpaul
  • 44,000
2

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 $$\log(A)=-1+\frac{\log (n)+\log (2 \pi )}{2 n}+\frac{1}{12 n^2}+O\left(\frac{1}{n^3}\right)$$

Claude Leibovici
  • 260,315