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$ \lim\limits_{n \to{+}\infty}{\sqrt[n]{n!}}$ is infinite

Please prove: $$ \lim_{n\to \infty}\sqrt[n]{\frac{1}{n!}} = 0 $$

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marked as duplicate by sdcvvc, Matt N., draks ..., rschwieb, Douglas S. Stones Oct 13 '12 at 1:26

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Have you tried using en.wikipedia.org/wiki/Stirlings_approximation ? –  Zach L. Oct 3 '12 at 12:48
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"Please do my homework." Well, at least you were polite... –  rschwieb Oct 3 '12 at 12:49
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Please show what you have tried or tell us what is causing trouble. This helps to address whatever you don't understand. –  robjohn Oct 3 '12 at 17:12
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Could someone who has upvoted this question explain why they did so? –  Noah Snyder Oct 3 '12 at 17:38
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@robjohn: Thank you for telling me this. I'll post more next time I ask. –  TheoYou Oct 4 '12 at 14:13

3 Answers 3

up vote 7 down vote accepted

$$0<\sqrt[n]{\frac{1}{n!}}=\left(1\cdot\frac{1}{2}\cdots\frac{1}{n}\right)^{\frac{1}{n}}\leq\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}<\frac{1+\ln n}{n}$$

As $\lim_{n\rightarrow\infty}\frac{1+\ln n}{n}=0$, so $\lim_{n\rightarrow\infty}\sqrt[n]{\frac{1}{n!}}=0$

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Squeeze theorem and aritmetic-mean-geometric-mean inequality used. en.wikipedia.org/wiki/Squeeze_theorem en.wikipedia.org/wiki/Arithmetic-geometric_mean_inequality –  mick Oct 3 '12 at 15:50
    
@mick Also $$\sum_{k=2}^n\frac{1}{k}<\log n$$ –  Pedro Tamaroff Oct 6 '12 at 16:25
    
Right Peter Tamaroff. –  mick Oct 6 '12 at 22:13

Hint: When writing out $n!$, you have \[ n! = n \cdot (n-1) \cdots \left\lceil \frac n2\right\rceil \cdots 1 \] so at least $\lfloor \frac n2\rfloor$ of the factors are larger then $\lceil \frac n2\rceil$. So $n! \ge \lceil \frac n2\rceil^{\lfloor \frac n2\rfloor}$.

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Just try to put n equal to infinity. since 'n' appears in the denominator it will tend to zero.

Please confirm the answer.

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(Downvoters, be nice.) Since you're new to the site, I suggest you take a look at other answers to see what people upvote. In the general case, it is better if you prove your claim, or at least sketch a proof. In this case, the problem is you get an indeterminate form $$\infty^0$$ Do you see why? –  Pedro Tamaroff Oct 6 '12 at 16:34
    
@fondoflior: No, you cannot deal with it like that. Because the degree is $\frac{1}{n}$ tend to zero too. If your method is feasible, how about $\sqrt[n]\frac{1}{n}$? –  Alfred Chern Oct 6 '12 at 16:36

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