Convergence: $\sum\frac1{n\ln(n^3)}$

Question

How do you test the convergence of

$$\sum_{n=2}^\infty\frac1{n\ln(n^3)}$$

I tried using limit comparison test, but had no conclusion.

yes, I mean to take the series. I don't know how to put the symbol sorry — Lazy Frog, Nov 26 '13 at 22:59
@Isaac: What was the point of changing the correct $\ln$ to the less readable $ln$ and making a meaningless change in the coding for the fraction? — Brian M. Scott, Nov 26 '13 at 23:05
Well how does $\text{log}(\text{log}(x))$ behave as $x \rightarrow \infty$? — Christopher K, Nov 26 '13 at 23:12
@BrianM.Scott: I thought this was a maths site, not a typographical inquisition. My edit may have been pointless, yet I think it contributed considerably more than the comment you left below it. — Isaac, Nov 26 '13 at 23:26
@Isaac: It defaced the previous edit and added nothing positive. It appears to have been made simply to make an edit. — Brian M. Scott, Nov 26 '13 at 23:30
Possible duplicate of Convergence of $\sum\limits_{n=2}^\infty \frac{1}{n^\alpha \ln^\beta (n)} $ for nonnegative $\alpha$ and $\beta$ — Nosrati, Nov 03 '18 at 06:45

score 4 · Answer 1 · answered Nov 26 '13 at 22:59

4

If you mean the series $$\sum_{n\ge 2}\frac1{n\ln n^3}\;,$$ note that $$\frac1{n\ln n^3}=\frac1{3n\ln n}\;,$$ and use the integral test.

answered Nov 26 '13 at 22:59

Brian M. Scott

score 0 · Answer 2 · answered Nov 27 '13 at 02:40

0

Use Cauchy's test, that is $\sum_{n \geq 2} \frac{2^n}{2^n \ln(2^{3n})} = \sum_{n \geq 2} \frac{1}{3n}$

The RHS diverges.

answered Nov 27 '13 at 02:40

Lemon

2 Answers2