Hey I've been having problem finding the limit of the sequence below
$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}$
Thanks!
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See this post. – David Mitra Nov 23 '13 at 13:15
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Or this one : https://math.stackexchange.com/questions/201906/showing-that-frac-sqrtnnn-rightarrow-frac1e – Arnaud D. Oct 24 '18 at 13:30
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The solution I know involves the p-norms and the Gamma function: in fact let $f: [0, \infty[ \to \mathbb{R}$ be defined by $f(x)=xe^{-x}$. Since $f$ is positive, continuous, and $0=f(0)=f(\infty)$, we have $\|f\|_\infty = \max f([0,\infty])$; now $f'(x)=(1-x)e^{-x}$ is $0$ for $x=1$, so that $\|f\|_\infty = f(1)=1/e$. Moreover $$\int_0^\infty |f(x)|^p \ dx=\int_0^\infty x^p e^{-px} dx = \int_0^\infty \frac{t^p}{p^p} e^{-t} \frac{dt}{p} = \frac{\Gamma(p+1)}{p^{p+1}}$$ then $$\|f\|_p = \frac{(\Gamma(p+1))^{1/p}}{p p^{1/p}}$$ since $\lim_{p \to \infty} \|f\|_p = \|f\|_{\infty} = 1/e$ and $\lim_{p \to \infty} p^{1/p}=1$, we get $$\lim_{p \to \infty} \frac{(\Gamma(p+1))^{1/p}}{p p^{1/p}} = \frac{1}{e} \Longrightarrow \lim_{n \to \infty} \frac{(n!)^{1/n}}{n}=\frac{1}{e} $$
I hope that this will help.
gangrene
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I've used a proposition proved on the notes of my professor:
if $|f|\infty = \infty$ or if $f \in L^p(\mu)$ for some $p>0$ then $|f|\infty = \lim_q |f|_q$.
– gangrene Nov 23 '13 at 13:12
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Use the simplest approximation for (n!) that is to say Log[n!] = n Log[n] - n
http://en.wikipedia.org/wiki/Stirling%27s_approximation.
Taking logarithms, you will easily show that the limit is 1/e as shown by Dedalus.
Claude Leibovici
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