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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?

Start wearing purple
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Andy
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3 Answers3

14

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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5

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

robjohn
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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.

Mikhail Katz
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  • 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. – Start wearing purple Oct 26 '13 at 17:10
  • 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. – Mikhail Katz Oct 26 '13 at 17:30
  • 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. – Start wearing purple Oct 26 '13 at 17:38
  • 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). – Mikhail Katz Oct 26 '13 at 17:48
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    Sure, but this is a well-known formula. Actually one also has explicit all-order version, see e.g. formulas (5)-(7) here. – Start wearing purple Oct 26 '13 at 17:52
  • @user72694: You don't need anything more than Stirling's asymptotic formula to get O.L.'s answer. In fact, using only the asymptotic formula you can get more. If you use the asymptotic series, you can get much more. – robjohn Oct 27 '13 at 15:45
  • @robjohn, thanks, that looks fine. Also O.L.'s corrected solution is fine. – Mikhail Katz Oct 27 '13 at 16:13