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Possible Duplicate:
Finding the limit of $\frac {n}{\sqrt[n]{n!}}$

Evaluate $\lim_{n \to \infty }\frac{(n!)^{1/n}}{n}.$

Can anyone help me with this? I have no idea how to start with. Thank you.

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    Use logs. Answer is 1/e2012-08-04

3 Answers 3

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Let's work it out elementarily by wisely applying Cauchy-d'Alembert criterion:

$\lim_{n\to\infty} \frac{n!^{\frac{1}{n}}}{n}=\lim_{n\to\infty}\left(\frac{n!}{n^n}\right)^{\frac{1}{n}} = \lim_{n\to\infty} \frac{(n+1)!}{(n+1)^{(n+1)}}\cdot \frac{n^{n}}{n!} = \lim_{n\to\infty} \frac{n^{n}}{(n+1)^{n}} =\lim_{n\to\infty} \frac{1}{\left(1+\frac{1}{n}\right)^{n}}=\frac{1}{e}. $

Also notice that by applying Stolz–Cesàro theorem you get the celebre limit:

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

The sequence $L_{n} = (n+1)!^{\frac{1}{n+1}} - (n)!^{\frac{1}{n}}$ is called Lalescu sequence, after the name of a great Romanian mathematician, Traian Lalescu.

Q.E.D.

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    @James Perhaps to be more precise, one should say that this equality holds provided the limit $\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}$ exists. In the other words, the existence of $\lim\limits_{n\to\infty} \frac{a_{n+1}}{a_n}$ implies the existence of $\lim\limits_{n\to\infty} \sqrt[n]{a_n}$. More details can be found in the duplicate thread.2012-08-04
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We can use Stirling's Approximation for the factorial:

$n!\sim\sqrt{2\pi n}\left(\frac{n}{{\rm e}}\right)^{n}$

Therefore, your expression becomes:

$\lim_{n\to\infty}{\left(\frac{1}{n}\left(\sqrt{2\pi n}\left(\frac{n}{\rm e}\right)^{n}\right)^{\frac{1}{n}}\right)}=\lim_{n\to\infty}{\left(\frac{1}{n}\frac{n}{\rm e}\sqrt[n]{\sqrt{2\pi n}}\right)}$

We know that $\lim_{n\to\infty}{\sqrt[n]{an}}=1$, so we have:

$\lim_{n\to\infty}{\left(\frac{1}{\rm e}\sqrt[n]{\sqrt{2\pi n}}\right)}=\frac{1}{\rm e}$

Hope this helps!

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    @jasoncube: as long as the limit $x$ exists, $\log x=\lim (1/n)\log(an)=0$, then exponentiate to get 1.2012-08-04
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With an integral test for convergence: $\displaystyle \int_1^n \ln(x)dx \leq \sum\limits_{k=2}^n \ln(k) = \ln(n!) \leq \int_2^{n+1} \ln(x)dx$.

You can deduce that $\ln(n!)=n\ln(n)-n + o(n)$. So $\displaystyle \lim\limits_{n\to + \infty} \frac{(n!)^{1/n}}{n}= e^{-1}$.