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

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

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

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Have you tried using en.wikipedia.org/wiki/Stirlings_approximation ? – Zach L. Oct 3 '12 at 12:48
"Please do my homework." Well, at least you were polite... – rschwieb Oct 3 '12 at 12:49
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
Could someone who has upvoted this question explain why they did so? – Noah Snyder Oct 3 '12 at 17:38
@robjohn: Thank you for telling me this. I'll post more next time I ask. – TheoYou Oct 4 '12 at 14:13

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

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

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 $$\sum_{k=2}^n\frac{1}{k}<\log n$$ – Peter 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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Suppose it is not true. Then we have $\displaystyle\lim_{n\to\infty}\sqrt[n]{\frac{1}{n!}}=\frac{1}{x}$ for some $x\geq 1$

By the hypothesis, the limit is not $0$, so we can write it like this: $\displaystyle\lim_{n\to\infty}\sqrt[n]{n!}=x$. Thus, as $n$ goes to infinity, then $n!\to x^n$.

This is clearly a contradiction, because for all $n>x$ the left side ($n!$) increases far more then the right side ($x^n$).

EDIT: i now see that a similar proof has been posted.

EDIT 2: I have'nt shown yet that the limit does exist.

I'll show that the function $f(n)=\sqrt[n]{\frac{1}{n!}}$ is strictly descending:

We have that $n!<(n+1)^n$. (because $n>0$)

Multiplicate bot sides with $(n!)^n$ and we have $(n!)^{n+1}<((n+1)!)^n$. Take the $n*(n+1)$-th root of both sides gives $\sqrt[n]{n!}<\sqrt[n+1]{(n+1)!}$.

Thus, $\sqrt[n]{\frac{1}{n!}}>\sqrt[n+1]{\frac{1}{(n+1)!}}$.

Now, because $f(n)>0$ and $f$ is strictly descending, the function converges and the limit exists.

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You also have to show that the limit exists, otherwise you cannot infer that the limit is indeed 0. – M Turgeon Oct 3 '12 at 14:13
That's right. I edited my answer. – barto Oct 3 '12 at 17:06

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

(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? – Peter 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