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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 $0$.

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  • $\begingroup$ you cannot split the limit like that $\endgroup$
    – Vasili
    Mar 16, 2018 at 13:14
  • $\begingroup$ 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. $\endgroup$
    – King Tut
    Mar 16, 2018 at 13:14
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    $\begingroup$ $1 = (1/n)+ (1/n)+\cdots + (1/n)$, so $\lim _{n \to \infty} 1 = \lim_{n \to \infty} \frac{1}{n} + \lim _{n \to \infty}\frac{1}{n} + \cdots = 0+0+\cdots = 0$. $\endgroup$ Mar 16, 2018 at 13:14
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    $\begingroup$ @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? $\endgroup$
    – user
    Mar 16, 2018 at 13:27
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    $\begingroup$ @m_t_ That was helpful. $\endgroup$
    – shiva
    Mar 16, 2018 at 15:56

1 Answer 1

0
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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

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

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  • $\begingroup$ I got it through this theorem, I didn't know about it. $\endgroup$
    – shiva
    Mar 16, 2018 at 15:55
  • $\begingroup$ @shiva it is a kind of Hopital for sequecences, very useful indeed, Bye $\endgroup$
    – user
    Mar 16, 2018 at 15:57

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