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I would like to show that $\frac{\log(n!)}{n\log n}$ increases as n increases, for positive integer n only (to ignore the use of gamma function). Note here $\log n = \ln(n)$, so using base e. I would like to be able to show this so I can then apply the Monotone Convergence Theorem to show that $(\log(n!)\sim(n\log n)$. My first idea was to try and show that the second derivative is always positive, but due to $\log(n!)$ not being continuous (without having to use gamma function which I feel is OTT for this question) this, of course, would not work. Does anyone have any ideas about how to prove that this sequence is increasing rigorously?

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Hint: You can use Stirling approximation for $n!$: $$n! \sim \sqrt{2\pi n}\left(\frac{n}{e}\right)^n$$ Or: $$n! = \sqrt{2\pi n}\left(\frac{n}{e}\right)^n \left(1+O\left(\frac{1}{n}\right)\right)$$ To know more about how to use this approximation you can see this link.

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  • $\begingroup$ This is not rigorous if one simply uses the approximation. $\endgroup$ – Carl Schildkraut Feb 28 '18 at 21:58
  • $\begingroup$ @CarlSchildkraut As he wants to approximate $\log(n!) \sim n \log n$, it can be helpful. $\endgroup$ – OmG Feb 28 '18 at 22:03
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    $\begingroup$ It can be helpful, but the approximation (without effective error terms) cannot act as a proof all on its own. The function $f(n)$ which equals $n!$ at $n\neq 10$ and $10^{10^{10}}$ at $n=10$ is asymptotic to $n!$, yet if $n!$ is replaced by $f(n)$ in the question, the relation is false as the term at $n=11$ is less than that at $n=10$. $\endgroup$ – Carl Schildkraut Feb 28 '18 at 22:09
  • $\begingroup$ @CarlSchildkraut you're right. More details about this approximation and bounds are provided in the inserted link. $\endgroup$ – OmG Feb 28 '18 at 22:13
  • $\begingroup$ @OmG Funnily enough I'm trying to prove this en route to learning about Stirling's formula, and so I am looking for a simpler proof without use of Stirling's approximation, which I am trying to derive via this! $\endgroup$ – Daniele1234 Feb 28 '18 at 23:28

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