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Theorem 3.29 in Walter Rudin's Principles of Mathematical Analysis, 3rd ed., states that

If $p>1$, then the series $$\sum_{n=2}^\infty \frac{1}{n (\log n)^p} $$ converges; if $p \leq 1$, the series diverges.

Now in the proof, Rudin only seems to discuss the case when $p> 0$, for it is only in this particular case that we can use the Cauchy Condensation Test.

How to deal with the case of $p<0$?

Of course, the case $p=0$ yields the divergent harmonic series.

Saaqib Mahmood
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  • Related : https://math.stackexchange.com/questions/9911/convergence-of-the-series-sum-limits-n-2-infty-frac1n-logs-n – Arnaud D. Dec 05 '19 at 10:40
  • Rudin relates this to theorem 3.28 ($\sum \frac{1}{n^p} $ converges if $p \gt 1$ and diverges, if $p \leq 1$). This was proved for all $p$. – Incompl33t Nov 09 '23 at 19:10

2 Answers2

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The case of $p<0$ is trivial, since $$\sum_{k=2}^\infty \frac{\ln^{-p} k}{k}\ge\sum_{k=4}^\infty \frac{1}{k}.$$

Hanul Jeon
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You can use the integral test:

$$ \sum_{n \geq 2 }\frac{1}{n ( \log n )^p} \cong \int\limits_2^{\infty} \frac{dz}{z (\log z )^p} = \int\limits_{2}^{\infty} \frac{ d(\log z)}{ ( \log z )^p }= ...$$

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    The integral test is not to be used; I'd like to do a a theorem in Rudin and so use the same armory that he's given me uptil this point. – Saaqib Mahmood Oct 25 '14 at 03:38