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Investigating the behavior of the following series:

$$\sum_{k=2}^\infty \frac{1}{\log^{p}k}$$

I broke it into 3 parts:

If $p = 0$ then it's just an infinite summation of ones, which diverges

If $p < 0$ then it diverges because log k goes to infinity as k goes to infinity

If $p > 0$ then it diverges by the integral test.

Is this right? Does it always diverge no matter what p is? Or did I do it wrong?

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  • $\begingroup$ It diverges for all $p$. Would have to see how you handled the integral test to know whether the computation was correct. $\endgroup$ – André Nicolas Apr 21 '13 at 0:23
  • $\begingroup$ @AndréNicolas I'm actually not super confident that I did the integral test right. I'm a little rusty on my integrations. Is there a better test to use for it? $\endgroup$ – Amber Apr 21 '13 at 0:30
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    $\begingroup$ Well, there is a fancy test called the Cauchy Condensation Test (look it up) that will do the job quickly. Integral test works OK, but you need to make estimates.And then you might as well compare with the harmonic, as suggested in one of the answers. In the long run, $n$ is (much) bigger than $(\ln n)^p$. This is almost familiar. Let's compare $x$ with $(\ln x)^p$. Let $y=\ln x$. Then we are comparing $e^y$ with $y^p$. The exponential wins. $\endgroup$ – André Nicolas Apr 21 '13 at 0:45
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Your work is just fine...very thorough, indeed. Try using the limit comparison test:

Compare $\log^p k\;$ with $\;\dfrac 1k$.

I.e., evaluate $\quad \lim_{k\to \infty} \dfrac{\log^p k}{k}.\;$

We get divergence if the limit evaluates to $0$, and it will do that whatever the value of $p$. (It diverges because the limit evaluates to $0$, and $\sum_{k\to \infty} \dfrac 1k$ diverges.)

Yes, your conclusion is correct: $$ \sum_{k=2}^\infty \frac{1}{\log^{p}k}$$ diverges no matter what the value of $p$.

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  • $\begingroup$ I like when OP show that level of work and it is easy guidance at that point! +1 $\endgroup$ – Amzoti Apr 21 '13 at 0:17
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You could also do it in one step, using the comparison (or limit comparison) test with the harmonic series.

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  • $\begingroup$ wouldn't it still depend on how big p is though? $\endgroup$ – Amber Apr 21 '13 at 0:34
  • $\begingroup$ With the comparison test, changing $p$ will change how many terms you need to ignore until $\log^p k < k$. With the limit comparison test, you'll get $\lim_{k\rightarrow \infty} \frac {\log^p k}{k}=0$, for all $p$. $\endgroup$ – vadim123 Apr 21 '13 at 0:53

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