# Interesting Series

1. Mar 9, 2013

### pierce15

1. The problem statement, all variables and given/known data

Is this series convergent or divergent?

$$\sum_{n=2}^{\infty} \frac{1}{(ln(n))^{ln \, ln(n)}}$$

2. Relevant equations

$$ln(x)<x^a, \, a>0$$

3. The attempt at a solution

I don't see a way to tackle this one other than the comparison test.

$$\sum_{n=2}^{\infty} \frac{1}{(ln(n))^{ln \, ln(n)}} < or > ?$$

What would I compare this to?

Last edited: Mar 9, 2013
2. Mar 9, 2013

### Ray Vickson

Re: LaTeX code: you need }} instead of } to close the \frac command properly. Also, you should write '\ln' instead of 'ln'; 'ln' produces $ln(n)$, while '\ln' gives $\ln(n)$.

3. Mar 9, 2013

### Zondrina

Slow down and take your hint into consideration.

For $a>0$, if $ln(n) < n^a$ then $(ln(n))^{ln(ln(n))} < (n^a)^{ln(ln(n))}$.

Reciprocating both sides, you get $\frac{1}{(ln(n))^{ln(ln(n))}} > \frac{1}{n^{aln(ln(n))}} >$ something I will allow you to notice for yourself.

Then you will have to break down the cases for positive a.... you'll need two cases in total. Pretty sure it was 'p-series' though, not 'a-series' :).

4. Mar 9, 2013

### I like Serena

Based on visual inspection it looks as if $\frac{1}{(\ln n)^{\ln(\ln n)}} > \frac 1 {n^{0.1}} > \frac 1 n$ for n sufficiently large.
The series diverges.

I don't see how I can match that with your observation though.
The series $\dfrac{1}{n^{a\ln(\ln n)}}$ converges for any a>0, meaning it won't give conclusive evidence.