For which values of $p$ does the serie $\sum\limits_{n=2}^\infty\frac{1}{n^p\ln(p)}$ converge? I'm trying to use the ratio test but I can't get a simple term in which use limit easily enough.
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$\begingroup$ Do you mean $\ln(n)$? If so, a place to start is to prove that if $\varepsilon > 0$ then $\lim_{n \to \infty} \frac{n^p \ln(n)}{n^{p+\varepsilon}} = 0$. Apply that for $p<1$. Use direct comparison for $p>1$. Then you have to do something else for $p=1$. $\endgroup$– IanCommented Aug 23, 2015 at 14:49
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$\begingroup$ @Ian sure, thanks. Let me try to use that. $\endgroup$– Jose PaterninaCommented Aug 23, 2015 at 14:50
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1 Answer
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As noted in a comment, you can settle the case for any $p \neq 1$ using comparison tests combined with the fact that the harmonic series diverges, and $\sum 1/n^p$ converges for $p > 1$. For $p = 1$, either use an integral comparison test, or if you note that $\ln n = (\log_2 n) / (\log_2 e)$ then you can work with the series $\sum_n 1/n \log_2 n$ and use the Cauchy condensation test to determine convergence.
