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Show:$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \frac{1}{e}$$

So I can expand the numerator by geometric mean. Letting $C_{n}=\left(\ln(a_{1})+...+\ln(a_{n})\right)/n$. Let the numerator be called $a_{n}$ and the denominator be $b_{n}$ Is there a way to use this statement so that I could force the original sequence into the form of $1/\left(1+\frac{1}{n}\right)^n$

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Stirling's approximation en.wikipedia.org/wiki/Stirling's_approximation –  user17762 Sep 24 '12 at 23:39
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@Marvis I suspect that Stirling's approximation would not be in the spirit of the question. –  MJD Sep 24 '12 at 23:41
    
Related: math.stackexchange.com/questions/28476/… –  Jonas Meyer Dec 14 '12 at 7:44
    
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4 Answers 4

up vote 13 down vote accepted

I would like to use the following lemma:

If $\lim_{n\to\infty}a_n=a$ and $a_n>0$ for all $n$, then we have $$ \lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}=a \tag{*} $$

Let $a_n=(1+\frac{1}{n})^n$, then $a_n>0$ for all $n$ and $\lim_{n\to\infty}a_n=e$. Applying ($*$) we have $$ \begin{align} e&=\lim_{n\to\infty}\sqrt[n]{a_1a_2\cdots a_n}\\ &=\lim_{n\to\infty}\sqrt[n]{(\frac{2}{1})^1(\frac{3}{2})^2\cdots(\frac{n+1}{n})^n}\\ &=\lim_{n\to\infty}\sqrt[n]{\frac{(n+1)^n}{n!}}\\&= \lim_{n\to\infty}\frac{n+1}{\sqrt[n]{n!}}=\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}}+\lim_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=\lim_{n\to\infty}\frac{n}{\sqrt[n]{n!}} \end{align} $$ where we can use (*) to show that $$ \lim_{n\to\infty}\frac{1}{\sqrt[n]{n!}}=0. $$ It follows that $$ \lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}=\frac{1}{e} $$

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Yes!! Thank you!! –  Edgar Aroutiounian Sep 25 '12 at 0:39
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I cannot follow what you are doing in your 3rd equal sign. –  Martin Argerami Sep 25 '12 at 4:18
    
@MartinArgerami: Notice that $n!=1\cdot 2\cdots\cdot n$. –  Jack Sep 25 '12 at 15:01
    
Yes, now I see it. I though it was a limit manipulation and not just algebra. Thanks. –  Martin Argerami Sep 25 '12 at 15:03
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Have not found a way to rewrite your expression to get the desired result. However, here is a suggested approach.

Maybe rewrite the left-hand side as $$\sqrt[n]{\frac{n!}{n^n}}.$$

Take the logarithm. We get $$\frac{1}{n}\left(\log\left(\frac{1}{n}\right)+ \log\left(\frac{2}{n}\right)+\log\left(\frac{3}{n}\right)+\cdots+\log\left(\frac{n}{n}\right)\right).$$

Now think of the above sum as a Riemann sum for the not quite proper integral $$\int_0^1 \log x\,dx.$$

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This is really neat! –  Manny Reyes Sep 25 '12 at 0:01
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If $a_n \geq 0$, then the following inequality holds:

$$ \liminf_{n\to\infty} \frac{a_{n+1}}{a_n} \leq \liminf_{n\to\infty} \sqrt[n]{a_n} \leq \limsup_{n\to\infty} \sqrt[n]{a_n} \leq \limsup_{n\to\infty} \frac{a_{n+1}}{a_n}. $$

Now let $ a_n = n! / n^n $. Then it follows that

$$ \frac{a_{n+1}}{a_n} = \frac{\frac{(n+1)!}{(n+1)^{n+1}}}{\frac{n!}{n^n}} = \frac{1}{\left(1+\frac{1}{n}\right)^n},$$

and hence

$$ \liminf_{n\to\infty} \frac{a_{n+1}}{a_n} = \limsup_{n\to\infty} \frac{a_{n+1}}{a_n} = \frac{1}{e}. $$

This proves that $\sqrt[n]{a_n} \to e^{-1}$ .

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For the inequality used in this answer see this post. (And also other posts listed there among linked questions.) –  Martin Sleziak Jun 24 at 7:56
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It's straightforward if you use Cesaro-Stolz theorem and then the celebre Lalescu's limit.

$$\lim_{n\to\infty}\frac{\sqrt[n]{n!}}{n}= \lim_{n\to\infty} \sqrt[n+1]{(n+1)!}-\sqrt[n]{(n)!}=\frac{1}{e}.$$

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