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? – Pedro Oct 17 '12 at 23:34
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1From 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 $\log\log m$. – joriki Oct 17 '12 at 23:37
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Is this a yandex school problem? – Norbert Oct 17 '12 at 23:37
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@Norbert: What's a yandex school problem? – joriki Oct 17 '12 at 23:39
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@Norbert, yes, this's task from homework. – aam Oct 17 '12 at 23:53
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@joriki, thanks with help. really got confused and didn't see this solution. – aam Oct 18 '12 at 00:03
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@WickedSpirit now you can answer your question. – Norbert Oct 18 '12 at 00:17
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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. – Gerry Myerson Oct 18 '12 at 01:37
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I posted below answer to close this question. @joriki, Could you look and check it, please? – aam Oct 19 '12 at 00:12
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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 :-) – joriki Oct 19 '12 at 07:58
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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'... – Steven Stadnicki Oct 19 '12 at 00:24