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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.

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

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  • $\begingroup$ Stirling formula. $\endgroup$ – Will M. Nov 3 '18 at 3:49
  • $\begingroup$ 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 $n$. $\endgroup$ – achille hui Nov 3 '18 at 3:50
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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.

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