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I'm trying to calculate: $$\lim_{n\to\infty}\frac{n!}{n^n}$$ I'd guess it's $1$ as the degree of the denominator and numerator would be $n$ and their coefficients would be $1$ as well.

If it's $1$, then: $$\sum_{n = 1}^\infty \frac{n!}{n^n}$$ Would be divergent from the divergence test. The issue with this is I know that this sum is convergent, but the limit still seems like it should be $1$.

Razin
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vishu
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    That is not quite the case. $$\lim_{n\to\infty}{n!\over n^n} = 0.$$ This suggests that $n^n \gg n!$. Can you see why? – Decaf-Math Dec 21 '16 at 07:53
  • Note that Stirling's Approximation says that $n!\approx\sqrt{2\pi}n^{n+1/2}e^{-n+1/(12n)}$. You can use this to help convince yourself of the limit. – Mark Schultz-Wu Dec 21 '16 at 07:57
  • By using @Mark Tip: If $y=\lim_{n\to\infty}\frac{n!}{n^n}$, then $$ y^{1/n}=\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{\sqrt[n]{n^n}} =\lim_{n\to\infty}\frac{\frac{n}{e}}{n}=\lim_{n\to\infty}\frac{1}{e} \Longrightarrow y=\lim_{n\to\infty}\frac{1}{e^n}=\frac{1}{\infty}=0 $$ – Amin235 Dec 21 '16 at 09:11

2 Answers2

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Consider the reciprocal sequence $$\frac{n^n}{n!}=\frac{n.n.\dotsb n}{n(n-1)\dotsb 3.2.1}=1\left(\frac{n}{n-1}\right)\left(\frac{n}{n-2}\right) \dotsb \frac{n}{1} \geq n.$$ Thus $\frac{n^n}{n!} \to \infty$, hence $\frac{n!}{n^n} \to 0$.

Anurag A
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    Very correct, but why do you consider the inverse ? –  Dec 21 '16 at 08:02
  • @YvesDaoust A very silly reason, while I was typing I had switched the numerator and denominator so I thought instead of cut paste, I should change my reasoning :-) – Anurag A Dec 21 '16 at 08:04
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In the expansion of $n!$, half of the factors do not exceed $n/2$, while the others do not exceed $n$. So when dividing by $n^n$, the ratio does not exceed $1/2^{n/2}$and both the sequence and the series converge.