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It is really not obvious to me which values of $p$ make $\sum_2^\infty \frac{1}{n^{p}log(n)}$ converge only using comparison test (general comparison test, limit comparison test).

Am I missing something ? I knew that the series converges for $p > 1$ and diverges for $p \leq 0$ by comparing it with $1/n^{p}$ but that's it.

Edit. I know it can be solved using integral test, cauchy condensation test, etc. but I am just wondering if there is a comparison to some other series that directly results in divergence of this one for $p \leq 1$ (or just $ p < 1$ is fine, too.)

Geralt of Rivia
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1 Answers1

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For $p<1$ compare with $\sum 1/n^q$ with $p<q<1$ and use the fact that for all $\epsilon>0$ $$ \lim_{n\to\infty}\frac{\log n}{n^\epsilon}=0. $$

Julián Aguirre
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