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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? – Pedro Tamaroff Oct 17 '12 at 23:34
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 $\log\log m$. – joriki Oct 17 '12 at 23:37
Is this a yandex school problem? – Norbert Oct 17 '12 at 23:37
@Norbert: What's a yandex school problem? – joriki Oct 17 '12 at 23:39
@Norbert, yes, this's task from homework. – aam Oct 17 '12 at 23:53
up vote 2 down vote accepted

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 0:24

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