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The task is to find asymptotic behavior of sum: $$\sum\limits_{k=2}^{m}\frac{1}{\ln(k!)}$$ when $m\to\infty$.

Any help with solving this one?

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    Can you use Stirling's approximation?2012-10-17
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    From Stirling's approximation, the denominator goes like $k\log k - k$, so I'd expect the sum to go like $\sum_k1/(k\log k)$, and since the integral of $1/(x\log x)$ is $\log\log x$, that should be asymptotic to $\log\log m$.2012-10-17
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    Is this a yandex school problem?2012-10-17
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    @Norbert: What's a yandex school problem?2012-10-17
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    @Norbert, yes, this's task from homework.2012-10-17
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    @joriki, thanks with help. really got confused and didn't see this solution.2012-10-18
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    @WickedSpirit now you can answer your question.2012-10-18
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    I believe Norbert is suggesting you can post an answer here and then, if it holds up, you can accept it. That way we don't have lots of seemingly-but-not-really unanswered questions lying around.2012-10-18
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    I posted below answer to close this question. @joriki, Could you look and check it, please?2012-10-19
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    @WickedSpirit: It's not quite rigorous, since you don't justify dropping the $O(n)$ term or replacing the sum by an integral, but I don't know what level of rigour you were aiming for -- it's good enough to convince me :-)2012-10-19

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Using Stirling's approximation: $$\ln(n!)\sim n\ln(n)+O(n)$$

Next we approximate sum with integral: $$\sum\limits_{k=2}^{m}\frac{1}{k\ln(k)}\sim\int_{2}^{m}\frac{dx}{x\ln(x)}=\ln\ln(m)-\ln\ln(2)$$

Found asymptotic behavior — $\ln \ln(n)$.

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    Looks good, although it's probably worth justifying the approximation of the sum with an integral - even if only by saying 'by Euler-MacLaurin'...2012-10-19