lframConvergence of $\sum_{n=2}^\infty \frac1{\log(n!)}$ [duplicate]

This question already has an answer here:

How do I show that this sum diverges/converges? $$\sum_{n=2}^\infty \frac1{\log(n!)}$$

I want to use the comparison test, but I do not know how to approach.

Also, Wolfram says this diverges by the comparison test, but Mathematica gives me a numerical answer using the NSum function.

marked as duplicate by Nosrati, Lord Shark the Unknown, ArsenBerk, Namaste, Parcly TaxelNov 3 '18 at 10:21

• Since $\log n! \sim n\log n$ for large $n$, the sum diverges like $\int_?^n \frac{dx}{x\log x} \sim \log\log n$ for large $n$. – achille hui Nov 3 '18 at 3:50
Hint: We have $$\log(n!)=\log n+\log(n-1)+\cdots+\log2<\log n+\log n+\cdots\log n Now apply the comparison test and make use of the integral test.