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I am struggling in this limit: $$\lim_{n\to \infty} \frac{n!}{n^n}$$ I really worked on this for few days: clearly, it goes to zero but I couln't do it rigorously. I'd really appreciate your help. Thanks from now.

Bora
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    Use Stirling's formula. – Kavi Rama Murthy Dec 31 '20 at 08:03
  • Yeah, but I do not know it. My reference has put it an exercise after couple pages from limit definition. He proves with sandwich theorem but also says that reader must try to prove it his own. And I am trying xD. – Bora Dec 31 '20 at 08:10
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    Or just observe, $n!/n^n = \frac{n}{n}\cdot\frac{n-1}{n}\cdots \frac{2}{n} \cdot \frac{1}{n}$. The leading terms are all less than or equal to one so, $\frac{n!}{n^n}<\frac{1}{n}$ and the conclusion follows. – WA Don Dec 31 '20 at 08:11
  • @BrianMoehring then I should go for sandwich theorem I suppose thanks – Bora Dec 31 '20 at 08:13
  • Yes, and I erased my comment because WADon's comment is the same as mine, but with more detail. It's probably the simplest method. – Brian Moehring Dec 31 '20 at 08:15

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You may use the quotient criterion for series:

$$a_n = \frac{n!}{n^n}\Rightarrow \frac{a_{n+1}}{a_n}=\frac 1{\left(1+\frac 1n\right)^n}\stackrel{n\to\infty}{\longrightarrow}\frac 1e<1$$

Hence, the series $\sum a_n$ is convergent and hence $\lim_{n\to\infty} a_n = 0$.

trancelocation
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A quick and dirty way to show it goes to 0 really fast: $$ n! \leq n^{\lfloor n/2\rfloor}\cdot (n/2)^{\lceil n/2\rceil} $$ So $$ \frac{n!}{n^n} \leq \frac{1}{2^{\lceil n/2\rceil}} $$

Clement C.
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$\begin{aligned} y=& \lim _{n \rightarrow \infty} \frac{n !}{n^{n}} \\ \ln y &=\lim _{n \rightarrow \infty} \ln \frac{n !}{n^{n}} \\ &=\lim _{n \rightarrow \infty}(n \ln n-n-n \ln n) \\ &=\lim _{n \rightarrow \infty}-n \\ &\\ \ln y \rightarrow-\infty \\ \Rightarrow & y \rightarrow 0 \end{aligned}$

ConfusedHuman
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