# Convergence of a logaritmic series

1. Oct 24, 2013

### Rono

1. The problem statement, all variables and given/known data
Analyze the convergence of the following series, describing the criteria used:
$\displaystyle\sum_{n=9}^{\infty}\frac{1}{(ln(ln(n)))^{ln(n)}}$

2. Relevant equations
None

3. The attempt at a solution
Wolfram Alpha says it converges due to comparison test, however I can't find to get a proper comparison. My main attempt was starting with $\displaystyle ln(n)< n$ and, starting from there, getting:
$(ln(ln(n)))^{ln(n)} < (ln(n))^{ln(n)} < n^{ln(n)} < n^{n}$
However, after getting their reciprocal, I manage to prove $\frac{1}{n^{n}}$ converges, but that is inconclusive. Any idea to which function I should start with to get the comparison test right?

2. Oct 24, 2013

### Office_Shredder

Staff Emeritus
If you want to show it converges, then you need to replace $\ln(\ln(n))^{\ln(n)}$ with something that is smaller than it, not larger. That means you need to think of a function which is smaller than ln(n) to replace some of those logs with.

It might help to remember that whenever there's a ln(n) in the exponent, you can do some algebra on it to rewrite your function!

3. Oct 25, 2013

### Rono

All right, managed to get $\displaystyle ln(ln(n))^{ln(n)} = n^{ln(ln(ln(n)))}$. Guess using $n^{2}$ as the comparison function would work, right? Because at infinity, $n^{ln(ln(ln(n)))} ≈ n^{n}$, therefore I would it should converge.

4. Oct 25, 2013

### Office_Shredder

Staff Emeritus
I don't think that it looks anything close to nn but I agree that n2 is a good comparison.