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

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

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

Have you tried using http://en.wikipedia.org/wiki/Stirlings_approximation ? — Zach L., Oct 03 '12 at 12:48
Please show what you have tried or tell us what is causing trouble. This helps to address whatever you don't understand. — robjohn, Oct 03 '12 at 17:12
Could someone who has upvoted this question explain why they did so? — Noah Snyder, Oct 03 '12 at 17:38
@robjohn: Thank you for telling me this. I'll post more next time I ask. — Qian, Oct 04 '12 at 14:13

score 9 · Answer 1 · answered Oct 03 '12 at 12:55

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 $n! \ge \lceil \frac n2\rceil^{\lfloor \frac n2\rfloor}$.

score 7 · Accepted Answer · edited Oct 04 '12 at 16:31

7

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

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

edited Oct 04 '12 at 16:31

robjohn

345,667

answered Oct 03 '12 at 12:59

Alfred Chern

2,658

Squeeze theorem and aritmetic-mean-geometric-mean inequality used. http://en.wikipedia.org/wiki/Squeeze_theorem
http://en.wikipedia.org/wiki/Arithmetic-geometric_mean_inequality
– mick Oct 03 '12 at 15:50
@mick Also $$\sum_{k=2}^n\frac{1}{k}<\log n$$ – Pedro Oct 06 '12 at 16:25
Right Peter Tamaroff. – mick Oct 06 '12 at 22:13

score -3 · Answer 3 · answered Oct 06 '12 at 16:27

-3

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

Please confirm the answer.

answered Oct 06 '12 at 16:27

fondoflinux

1

2

(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 Oct 06 '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 06 '12 at 16:36

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

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