TITLE: Theorem 3.29 in Baby Rudin
QUESTION [5 upvotes]: 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.

REPLY [0 votes]: 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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