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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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marked as duplicate by Martin Sleziak, Did, J. M. is back., t.b., David Mitra Aug 4 '12 at 14:41

This question has been asked before and already has an answer. If those answers do not fully address your question, please ask a new question.

Use logs. Answer is 1/e – i. m. soloveichik Aug 4 '12 at 13:41

3 Answers 3

up vote 6 down vote accepted

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.


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For those, like me, who didn't know what the "Cauchy-d'Alembert criterion" was, it seemingly says that $$\lim_{n \rightarrow \infty} (a_n)^{1/n} = \lim_{n \rightarrow \infty} \frac{a_{n+1}}{a_n}$$ – James Fennell Aug 4 '12 at 14:25
@ James Fennell: that's correct! – Chris's sis the artist Aug 4 '12 at 14:37
@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. – Martin Sleziak Aug 4 '12 at 15:06

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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not $\frac{n}{2}$ but $\frac{n}{e}$ – Seyhmus Güngören Aug 4 '12 at 13:45
@SeyhmusGüngören Sorry typo'd, thanks! – Shaktal Aug 4 '12 at 13:47
Can you show me why $\lim_{n \to \infty }(an)^{1/n}=1$? – ᴊ ᴀ s ᴏ ɴ Aug 4 '12 at 13:54
@jasoncube: as long as the limit $x$ exists, $\log x=\lim (1/n)\log(an)=0$, then exponentiate to get 1. – Kevin Carlson Aug 4 '12 at 14:00

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 $\displaystyle \lim\limits_{n\to + \infty} \frac{(n!)^{1/n}}{n}= e^{-1}$.

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