# Asymptotic behavior of $\sum\limits_{k=2}^{m}\frac{1}{\ln(k!)}$

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?

• Can you use Stirling's approximation? 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 $\log\log m$. Oct 17 '12 at 23:37
• Is this a yandex school problem? Oct 17 '12 at 23:37
• @Norbert: What's a yandex school problem? Oct 17 '12 at 23:39
• @Norbert, yes, this's task from homework.
– 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 — $\ln \ln(n)$.