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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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    @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