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I want to calculate $\lim_{n\to\infty}\frac{\log n!}{n\log n}$. I know this is a duplicate and I read its equal to $1$, however I can't seem to find the problem in the below calculation:

$$\lim_{n\to\infty}\frac{\log n!}{n\log n} = \lim_{n\to\infty} \frac{\log1 +\log2+\dots+\log n}{n\log n}$$

Now if we split the limits, we get:

$$\lim_{n\to\infty} \frac{\log1}{n\log n}+\dots+\lim_{n\to\infty} \frac{\log n}{n\log n}$$

Then each term will give $0$, and thus answer should be $0$.

shiva
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  • you cannot split the limit like that – Vasili Mar 16 '18 at 13:14
  • Simple answer: Each term has something to contribute to the total expression, but this series has infinite number of such terms. If there were finite number of terms then there was no problem! Also United we stand divided we fall. – King Tut Mar 16 '18 at 13:14
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    $1 = (1/n)+ (1/n)+\cdots + (1/n)$, so $\lim {n \to \infty} 1 = \lim{n \to \infty} \frac{1}{n} + \lim _{n \to \infty}\frac{1}{n} + \cdots = 0+0+\cdots = 0$. – Matthew Towers Mar 16 '18 at 13:14
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    @shiva The sum of infinitely many small term can lead to any number and also diverge, in this case the sum is 1, let try to continue the solution by Stolz-Cesaro, can you find the limit? – user Mar 16 '18 at 13:27
  • We know that: $$1 - \dfrac{1}{\log(n)} \leq \dfrac{\log(n!)}{n \log(n)} \leq 1$$ and when $n\to \infty$ what happens by squeez theorem? – Mehrdad Zandigohar Mar 16 '18 at 14:19
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    @m_t_ That was helpful. – shiva Mar 16 '18 at 15:56
  • @MehrdadZandigohar I will have to read about squeez theorem. Thanks. – shiva Mar 16 '18 at 15:58
  • @shiva It is like $1 \leq x \leq 1 \to x=1$ – Mehrdad Zandigohar Mar 16 '18 at 17:31
  • @MehrdadZandigohar How did you come up with $1 - \dfrac{1}{\log(n)} \leq \dfrac{\log(n!)}{n \log(n)}$? – shiva Mar 16 '18 at 19:14

1 Answers1

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HINT

This method is not valide since you are summing up infintely many terms.

As an alternative refer to Stolz-Cesaro

$$\lim_{n\to\infty}\frac{\log n!}{n\log n} = \lim_{n\to\infty}\frac{\log (n+1)!-\log n!}{(n+1)\log (n+1)-n\log n}$$

user
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