I have to proof the following:
$ \lim\limits_ {n\to\infty} \dfrac{\sqrt[n]{n!}}{n} = \frac{1}{e}$
Do you have any hints for me, since I do not know where to start..
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Taking logarithm also helps. – Sangchul Lee May 07 '13 at 09:20
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1One way: It is known that for a sequence $(a_n)$ of positive terms, if $\lim_{n\rightarrow\infty} {a_{n+1}\over a_n}$ exists, then so does $\lim_{n\rightarrow\infty} \root n\of {a_n}$ and the two limits are equal (see Lemma 3 here). Apply this to $a_n=n!/n^n$ – David Mitra May 07 '13 at 10:18
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Here is a similar question. – David Mitra May 07 '13 at 10:21
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You will be interested in Stirling's Approximation.
$n! \sim (\dfrac{n}{e})^n \sqrt{2 \pi n}$
Inceptio
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$\frac{^n\sqrt{n!}}{n}$ = $\frac{^n\sqrt{(\frac{n}{e})^{n}\sqrt{2πn}}}{n}$ = $ \frac{1}{e}( ^n\sqrt{2πn}) $
Now I can use that $lim (^n\sqrt{n})= 1 $ and therefore the limit is $1/e $?
– Vazrael May 07 '13 at 09:29 -
@KevinLiebing: Yes, pretty fine. You have to use L'hopitals rule to find the limit of $^n\sqrt{n}$ – Inceptio May 07 '13 at 09:37