My teacher gave me this limit to solve. She said it was meant to be really hard and she was right. I have no idea on how to attack this problem. Would anyone be able to help?
Question:
Prove that $$\lim_{n\to\infty} (\frac{n!}{n^{n}})^{\frac{1}{n}}=\frac{1}{e} $$
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$\begingroup$We can use the fact that if $a_n>0$ for all $n\ge1$ and the sequence $\frac{a_{n+1}}{a_n}$ converges in $[0,\infty]$, then $$ \lim_{n\to\infty}a_n^{1/n}=\lim_{n\to\infty}\frac{a_{n+1}}{a_n} $$ (see here).
We have that $$ \frac{(n+1)!}{(n+1)^{n+1}}\cdot\frac{n^n}{n!}=\biggl(1+\frac1n\biggr)^{-n}\to\frac1 e $$ as $n\to\infty$. Hence, $$ \lim_{n\to\infty}\biggl(\frac{n!}{n^n}\biggr)^{1/n}=\frac1e. $$
$\endgroup$ 1 $\begingroup$Disregard Stirling's formula, Riemann sums are enough here.
$$\frac{1}{n}\left[\log(n!)-n\log(n)\right]=\frac{1}{n}\sum_{k=1}^{n}\log\left(\frac{k}{n}\right)\to\int_{0}^{1}\log(x)\,dx = -1$$ hence by exponentiating both sides $$ \lim_{n\to +\infty}\left(\frac{n!}{n^n}\right)^{1/n} = \frac{1}{e} $$sic et simpliciter.
$\endgroup$ $\begingroup$There is a canonical way to approach this answer, which is to use Stirling's formula for $n!$, which essentially states that $$ n! \approx \sqrt{2 \pi n} \left( \frac{n}{e} \right)^n.$$ Inserting Stirling's formula, this becomes a very ordinary and straightforward limit.
Morally, this question is to determine the rough order of growth of $n!$, which is the main goal of Stirling's formula. So you might interpret this as an excuse to determine how to prove and show Stirling's formula. This has been addressed multiple times on this site, or on the wikipedia page, or elsewhere.
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