series convergence with comparison test $\frac{1}{n\log(n)^p}$

Question

I came across a question about convergence tests and I'm not sure how to show that

$\sum_{n=2}^{\infty}\frac{1}{n \log(n)^p}$

converges for $p>1$. I found that the integral test works but I haven't learnt it yet. I have learnt the comparison test, so I tried that but couldn't find any series that converges to which I can compare.
So my question is both how can I show convergence here with comparison test and also: Is there some list of convergent series that I can compare to?

PS: I also found out that this is called a ln series

Also, this is not a duplicate of this question. (The power is on the $\log$ here)

You could apply https://en.wikipedia.org/wiki/Cauchy_condensation_test, which isn't very hard to prove. — PseudoNeo, Dec 23 '16 at 10:05
@ClementC. really? never heard of it. I'll search now. Thanks! — ItamarG3, Dec 23 '16 at 10:05
Possible duplicate of Convergence of the series $\sum \limits_{n=2}^{\infty} \frac{1}{n\log^s n}$ — Arnaud D., Dec 05 '19 at 10:38
See also: Convergence of $\sum\frac{1}{n(\ln n)^c}$ and Convergence of the series $\sum \limits_{n=2}^{\infty} \frac{1}{n\log^s n}$ — Martin Sleziak, Dec 21 '19 at 12:52

score 9 · Accepted Answer · answered Dec 23 '16 at 10:57

HINT Use the Cauchy Condensation Test: $\sum a_n$ converge iff $\sum 2^na_{2n}$ converge.

So $\sum \frac{1}{n \log(n)^p}$ converge iff $\sum 2^n\frac{1}{2^n \log(2^n)^p}$ converge.

While $\sum 2^n\frac{1}{2^n \log(2^n)^p}=\sum \frac{1}{(n\log 2)^p}=\frac{1}{(\log 2)^p}\sum\frac{1}{n^p}$

Paolo Leonetti · Answer 2 · 2016-12-23T10:20:51.300

6

$$ \sum_{n\ge 2}\frac{1}{n\log^p n}\sim \int_2^{\infty}\frac{1}{x\log^p x}\mathrm{d}x <\infty, $$ where the last $<$ is motivated by the fact that a primitive is $\frac{1}{\log^{p-1} x}$ (and $p-1>0$).

edited Dec 23 '16 at 10:20

answered Dec 23 '16 at 10:09

Paolo Leonetti

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As I've said, I haven't learnt the integral test, and I am trying the comparison test. – ItamarG3 Dec 23 '16 at 10:36
1

You can see it as a comparison test, since the LHS is smaller than the integral, up to constants. – Paolo Leonetti Dec 23 '16 at 12:54

series convergence with comparison test $\frac{1}{n\log(n)^p}$

2 Answers2

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