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For what values of $p$ is the series $\sum{\frac{1}{n(\log(n))^p}}$ divergent and for what values it is convergent?

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  • $\begingroup$ Divergent for $p\le 1$, convergent for $p\gt 1$. The Integral Test will work nicely. So will the likely less familiar Cauchy Condensation Test. $\endgroup$ Aug 2 '15 at 13:46
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This is a Bertrand's series. It converges for any real exponent $p>1$, and diverges for $p\le 1$. It trivially diverges if $p\le 0$. If $0<p\le 1$, it is enough to prove it diverges for $p=1$. As the function $\dfrac1{x\ln x}$ is continuous, decreasing for $x>\dfrac1e$, we may apply Cauchy's integral test: $$\int_2^{+\infty}\frac{\mathrm d\mkern1mu x}{x\ln x}=\ln\ln x\biggr\rvert_2^{+\infty}\enspace\text{diverges,}$$ hence the series $\displaystyle \sum_{n=2}^{\infty}\dfrac1{n\ln n}\;$ diverges.

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