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Evaluate

$$\lim_{n\to\infty}\frac{n}{\ln n}\left(\frac{\sqrt[n]{n!}}{n}-\frac{1}{e}\right).$$

This sequence looks extremely horrible and it makes me crazy. How can we evaluate this?

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Use Stirling's approximation: $$n!\sim\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}=\left(\frac{n}{e}\right)^{n}e^{\frac12 \ln 2\pi n}$$ It transforms your limit into $$\lim_{n\rightarrow\infty}\frac{n}{\ln n}\frac{e^{\frac{\ln n}{2n}}-1}{e}=\lim_{x\rightarrow 0}\frac{e^{\frac{x}{2}}-1}{ex}=\frac{1}{2e}.$$ When obtaining the first expression, we neglect $\frac{\ln 2\pi}{2n}$ in the exponential (it is $o\left(\frac{\ln n}{2n}\right)$), and then we make the change of variables $x=\frac{\ln n}{n}$.

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  • $\begingroup$ Quite elegant! Thanks! $\endgroup$ – Andy Oct 25 '13 at 3:18
  • $\begingroup$ Sorry, I am still a little confused about the change of variable you used, could you please specify? Thanks $\endgroup$ – Andy Oct 25 '13 at 3:45
  • $\begingroup$ @Andy Sure, I have edited my answer. $\endgroup$ – Start wearing purple Oct 25 '13 at 3:50
  • $\begingroup$ Now it is clear. Thanks $\endgroup$ – Andy Oct 25 '13 at 3:50
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    $\begingroup$ @OL: There is a problem with this proof because Stirling's formula does not give you an exact value of $n!$ but only an asymptotic approximation. So long as you work with multuplication the asymptotic behavior is sufficient, but if you with to subtract off the term $\frac{1}{e}$ as in your calculation, further justification is required... $\endgroup$ – Mikhail Katz Oct 25 '13 at 8:47
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Stirling's asymptotic formula (without any series enhancement) $$ n!\sim\frac{n^n}{e^n}\sqrt{2\pi n}\tag{1} $$ put into Landau notation is $$ n!=\frac{n^n}{e^n}\sqrt{2\pi n}\,(1+o(1))\tag{2} $$ which gives $$ \begin{align} (n!)^{1/n} &=\frac ne(2\pi n)^{\frac1{2n}}(1+o(1/n))\\ &=\frac ne\left(1+\frac{\log(2\pi n)}{2n}\right)(1+o(1/n))\\ \frac{(n!)^{1/n}}{n} &=\frac1e\left(1+\frac{\log(2\pi n)}{2n}\right)(1+o(1/n))\\ \frac{(n!)^{1/n}}{n}-\frac1e &=\frac1e\frac{\log(2\pi n)}{2n}+o(1/n)\\ \frac{n}{\log(n)}\left(\frac{(n!)^{1/n}}{n}-\frac1e\right) &=\frac1{2e}+\frac{\log(2\pi)}{2e\log(n)}+o(1/\log(n))\tag{3} \end{align} $$ Taking the limit as $n\to\infty$, $(3)$ yields $$ \lim_{n\to\infty}\frac{n}{\log(n)}\left(\frac{(n!)^{1/n}}{n}-\frac1e\right) =\frac1{2e}\tag{4} $$ which is the result sought.

However, $(3)$ gives more $$ \lim_{n\to\infty}\log(n)\left(\frac{n}{\log(n)}\left(\frac{(n!)^{1/n}}{n}-\frac1e\right)-\frac1{2e}\right)=\frac{\log(2\pi)}{2e}\tag{5} $$

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There is a gap in @O.L.'s solution as I explained in a comment below that solution. I don't have time now to provide a complete solution but the OP should be aware of this. I see that the wiki page provides a formula with an error term, but the $O(1/n)$ occurs inside an argument and is not stated as a theorem, nor is there a source for this. It is probably true but note also that we need the error estimate for $\frac{\sqrt[n]{n!}}{n}$ rather than for $n!$ itself.

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  • $\begingroup$ If you scroll the Wikipedia page I refer to, you will find that $$n!\sim\left(\frac{n}{e}\right)^{n}\sqrt{2\pi n}\left[1+O\left( \frac1n \right) \right]$$ That $O\left( \frac1n \right)$ easily justifies the argument. Anyhow your post should better be a comment. $\endgroup$ – Start wearing purple Oct 26 '13 at 17:10
  • $\begingroup$ I see that the $O(1/n)$ occurs inside an argument and is not stated as a theorem, nor is there a source for this. It is probably true but note also that we need the error estimate for $\frac{\sqrt[n]{n!}}{n}$ rather than for $n!$ itself. $\endgroup$ – Mikhail Katz Oct 26 '13 at 17:30
  • $\begingroup$ Well, $[1+ O(1/n) ]^{1/n} = 1 + O(1/n^2) $. Do you want me to prove such details? I definitely do not intend to write all of them. $\endgroup$ – Start wearing purple Oct 26 '13 at 17:38
  • $\begingroup$ Hi @O.L., thanks for your comment! I don't "want" you to provide any details! All work here is on voluntary basis. I just wanted to mention (with the OP in mind) that some details need to be filled in that may not be obvious to a novice. Personally I was not familiar with the $1+O(1/n)$ form of Stirling's estimate; I hope the person who wrote up the wiki page knows what he is doing. This is not always the case, you know (I can provide some examples). $\endgroup$ – Mikhail Katz Oct 26 '13 at 17:48
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    $\begingroup$ Sure, but this is a well-known formula. Actually one also has explicit all-order version, see e.g. formulas (5)-(7) here. $\endgroup$ – Start wearing purple Oct 26 '13 at 17:52

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