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

Question

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

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.

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 03 '18 at 03:50

Clayton · Answer 1 · 2018-11-03T03:54:59.107

4

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.

edited Nov 03 '18 at 03:54

answered Nov 03 '18 at 03:49

Clayton

24,751

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

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