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Problem

Show, using the integral test that $$\displaystyle\sum\limits_{n=2}^{\infty}\frac1{n(\ln(n))^p}$$ converges for $p>1 $ and diverges otherwise.

My progress

So far, I have

$$\begin{align} \int_2^\infty \frac1{n(\ln n)^p}\mathrm dn &= \lim\limits_{a\to\infty}\left.\left[ \frac{(\ln n)^{1-p}}{1-p} \right]\right|_2^a \\ &= \lim\limits_{a\to\infty}\left[ \frac{(\ln a)^{1-p}}{1-p} \right] - \frac{(\ln 2)^{1-p}}{1-p} \end{align}$$

But I don't see where to go from here.

I know from a previous conundrum that $\int_1^\infty \frac{1}{x^p}\mathrm dx$ converges and diverges for the same criteria on $p$. Is that useful here?

Thanks in advance for any help!

  • 1
    Not related to the integral test: If you have to show convergence for anything involving logarithms, the Cauchy condensation test is likely helpful. Here, it has: $$\frac1{2\ln^p(2)}\sum_{n=1}^\infty\frac1{n^p}<\sum_{n=2}^\infty\frac1{n\ln^p(n)}<\frac1{\ln^p(2)}\sum_{n=1}^\infty\frac1{n^p}$$Assuming the obvious $p>0$ so that the sequence is monotone. Now it's a simple p-series. :D – 2017-02-15
  • 0
    Possible duplicate of [Convergence of $\sum\limits_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)} $ for nonnegative $\alpha$ and $\beta$](https://math.stackexchange.com/questions/267697/convergence-of-sum-limits-n-2-infty-frac1n-alpha-ln-beta-n-for) – 2018-11-03

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You need to understand if the integral converges or diverges. The only interesting piece is $$ \lim_{a\to\infty} \frac{(\ln a)^{1-p}}{1-p} $$ So clearly as $a \to \infty$, you also have $\ln a \to \infty$, so what must be the necessary and sufficient conditions on $p$ for $\ln(a)^{1-p} < \infty$?