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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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    Have you tried using http://en.wikipedia.org/wiki/Stirlings_approximation ?2012-10-03
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    "Please do my homework." Well, at least you were polite...2012-10-03
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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.2012-10-03
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    Could someone who has upvoted this question explain why they did so?2012-10-03
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    @robjohn: Thank you for telling me this. I'll post more next time I ask.2012-10-04

3 Answers 3

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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?2012-10-06
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    @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 $\sqrt[n]\frac{1}{n}$?2012-10-06
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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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$$0<\sqrt[n]{\frac{1}{n!}}=\left(1\cdot\frac{1}{2}\cdots\frac{1}{n}\right)^{\frac{1}{n}}\leq\frac{1+\frac{1}{2}+\cdots+\frac{1}{n}}{n}<\frac{1+\ln n}{n}$$

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. http://en.wikipedia.org/wiki/Squeeze_theorem http://en.wikipedia.org/wiki/Arithmetic-geometric_mean_inequality2012-10-03
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    @mick Also $$\sum_{k=2}^n\frac{1}{k}<\log n$$2012-10-06
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    Right Peter Tamaroff.2012-10-06