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

Question

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

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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.

Answer 1

Hint: We have $$\log(n!)=\log n+\log(n-1)+\cdots+\log2<\log n+\log n+\cdots\log n<n\log n.$$ Now apply the comparison test and make use of the integral test.