Which grows at a faster rate, $\ln(n!)$ vs $\ln(n^n)$? How to solve such type of questions considering $n \rightarrow \infty$
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Hint: $\ln(n!) = n\ln(n)-n+O(n)$. – JMoravitz Oct 24 '18 at 05:17
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1Although $n^n\gg n!$, $\lim_{n\to\infty}\frac{\ln n!}{\ln n^n}=1$. – Kemono Chen Oct 24 '18 at 05:28
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@Komono Chen Could you please explain your solution? – Vishesh Sharma Oct 24 '18 at 05:45
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https://math.stackexchange.com/questions/674002/which-has-a-higher-order-of-growth-n-or-nn – Peter Szilas Oct 24 '18 at 06:09
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$\ln n^n = n\ln n = \ln n + \ln n + \cdots + \ln n> \ln 1 + \ln 2 + \ln 3 + \cdots + \ln n = \ln n! $
Nilotpal Sinha
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Thanks! But is this applicable even when $n \rightarrow \infty$? – Vishesh Sharma Oct 24 '18 at 05:44
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@VisheshSharma Yes in the sense that $\lim_{n\rightarrow\infty}[\ln(n^n) - \ln(n!)]= \infty$. No in the sense that $\lim_{n_\rightarrow\infty}[\ln(n^n)/\ln(n!)] = 1$. – eyeballfrog Oct 24 '18 at 05:47
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