TITLE: Asymptotic behavior of $\sum\limits_{k=2}^{m}\frac{1}{\ln(k!)}$
QUESTION [6 upvotes]: 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?

REPLY [3 votes]: 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 — $\ln \ln(n)$.