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:) I need some help with calculating the limit $$\lim_{n\to\infty}\dfrac{\ln(n!)}{n\ln(n)}.$$ I think that it's easy to see that we can find it using d'Alembert (square root) criterion, because, obviously, $\forall n\in\mathbb{N}_{\geqslant 2}:\phi_n>0$, where $\left(\phi_n\right)_{n\in\mathbb{N}_{\geqslant 2}},\,\phi_n:=\dfrac{\ln(n!)}{n\ln(n)}$, so I did the following observation: $$\dfrac{\ln(n!)}{n\ln(n)}=\dfrac{1}{n}\cdot\dfrac{\ln(n!)}{\ln(n)}=\dfrac{\dfrac{1}{n}\ln(n!)}{\ln(n)}=\dfrac{\ln(n!)^{\frac{1}{n}}}{\ln(n)}=\dfrac{\displaystyle\sqrt[n]{\ln(n!)}}{\ln(n)}=\displaystyle\sqrt[n]{\dfrac{\ln(n!)}{\ln^n(n)}}.$$ Ok. d'Alembert tells me that I should do the limit $$\ell:=\lim_{n\to\infty}\bigg\lvert\dfrac{x_{n+1}}{x_n}\bigg\lvert=\lim_{n\to\infty}\dfrac{x_{n+1}}{x_n},$$ where $\left(x_n\right)_{n\in\mathbb{N}_{\geqslant 2}},\,x_n:=\dfrac{\ln(n!)}{\ln^n(n)}$. We have, then: $$\ell=\lim_{n\to\infty}\dfrac{\ln[(n+1)!]}{\ln^{n+1}(n+1)}\cdot\dfrac{\ln^n(n)}{\ln(n!)}=\cdots=\lim_{n\to\infty}\left[\dfrac{\ln(n)}{\ln(n+1)}\right]^n\left[\dfrac{1}{\ln(n+1)}+\dfrac{1}{\ln(n!)}\right]$$ but what about the next?

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    $\begingroup$ I would use Stirling's approximation on the factorial in the numerator. $\endgroup$ – Ross Millikan Oct 22 '17 at 14:54
  • $\begingroup$ Use Cesaro-Stolz to get the answer $1$ almost immediately $\endgroup$ – Paramanand Singh Oct 22 '17 at 14:56
  • $\begingroup$ This limit been asked here several times, see for example math.stackexchange.com/questions/376988/… or math.stackexchange.com/questions/181682/… (but for some reason none of the answers there are using stirling approximation...). $\endgroup$ – Sil Oct 22 '17 at 14:59
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    $\begingroup$ Watch out: $$\frac{1}{n} \operatorname{ln} (n!) = \operatorname{ln} ( (n!)^{\frac{1}{n}} ) \neq ( \operatorname{ln} (n!) )^{\frac{1}{n}} $$ $\endgroup$ – frafour Oct 22 '17 at 15:02
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By the Stirling formula,

$$\frac{\log n!}{n\log n}\approx\frac{\log \sqrt{2\pi n}+n\log n-n}{n\log n}\to1$$


Alternatively, by monotonicity of the logarithm,

$$\log\lfloor x-1\rfloor\le\log x<\log\lfloor x\rfloor$$

Then integrating from $1$ to $n+1$,

$$\log n!=\sum_{k=1}^n\log n\le \left.x(\log x-1)\right|_1^n=n\log(n)-n+1<\sum_{k=1}^{n+1}\log n=\log n!+\log(n+1)$$ and the limit $1$ follows.

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  • $\begingroup$ Inaccuracies are expected in this development, but the asymptotic behavior is right. $\endgroup$ – Yves Daoust Oct 22 '17 at 15:11
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By Stolz theorem we obtain: $$\lim_{n\rightarrow+\infty}\frac{\ln{n!}}{n\ln{n}}=\lim_{n\rightarrow+\infty}\frac{\ln{n}}{n\ln{n}-(n-1)\ln(n-1)}=$$ $$=\lim_{n\rightarrow+\infty}\frac{\frac{1}{n}}{\ln{n}+1-\ln(n-1)-1}=\lim_{x\rightarrow0^+}\frac{-x}{\ln(1-x)}=1.$$

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