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?

  • $\begingroup$ Can you use Stirling's approximation? $\endgroup$
    – Pedro Tamaroff
    Oct 17 '12 at 23:34
  • 1
    $\begingroup$ 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 $\log\log m$. $\endgroup$
    – joriki
    Oct 17 '12 at 23:37
  • $\begingroup$ Is this a yandex school problem? $\endgroup$
    – Norbert
    Oct 17 '12 at 23:37
  • $\begingroup$ @Norbert: What's a yandex school problem? $\endgroup$
    – joriki
    Oct 17 '12 at 23:39
  • $\begingroup$ @Norbert, yes, this's task from homework. $\endgroup$
    – aam
    Oct 17 '12 at 23:53

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 — $\ln \ln(n)$.

  • $\begingroup$ Looks good, although it's probably worth justifying the approximation of the sum with an integral - even if only by saying 'by Euler-MacLaurin'... $\endgroup$ Oct 19 '12 at 0:24

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