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I want to know the series $\sum_2^\infty\cos( \log n)/n\log n$ converges or diverges. The log n in the bottom makes every my effort futile....Could anyone please help?

Keith
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  • But you also have an $n$ in the bottom, which means maybe the integral test with a $u$-substitution? – Matthew Leingang Aug 15 '19 at 15:30
  • then I cannot tackle with the $\cos( \log n)$...Could you please explain in more detail? I am just stuck and frustrated..... – Keith Aug 15 '19 at 15:31
  • The integral test doesn't work here because $cos(t)/t$ is not a positive valued function.... – Keith Aug 15 '19 at 15:32
  • Ah, you're right about $\cos(t)/t$. Hm... – Matthew Leingang Aug 15 '19 at 15:35
  • I came to think that if I can show the partial sums of $\cos(\log n)/n$ are bounded, I can use the Dirichlet test. – Keith Aug 15 '19 at 15:40
  • This should help https://math.stackexchange.com/questions/936526/does-sum-n-1-infty-frac-cos-lnnn-converge with the bounded part of your question – QC_QAOA Aug 15 '19 at 15:57
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    @Keith: Unlike what is stated in most textbooks, the integral test applies in that the series $\sum_{n=1}^\infty f(n)$ and the integral $\int_1^\infty f(x) , dx$ converge and diverge together if $\int_1^\infty |f'(x)| , dx < \infty$. This is proved here. – RRL Aug 15 '19 at 16:18
  • Thank you for your valuable information... – Keith Aug 15 '19 at 16:26

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