# Series 4.

1. Nov 18, 2007

### azatkgz

1. The problem statement, all variables and given/known data
Determine whether the series converges or diverges.

$$\sum_{n=3}^{\infty}\frac{1}{\left(\ln (\ln n)\right)^{\ln n}}$$

3. The attempt at a solution

wait I'll try to solve

Last edited: Nov 18, 2007
2. Nov 18, 2007

### azatkgz

I deleted wrong solution here.

Last edited: Nov 18, 2007
3. Nov 18, 2007

### azatkgz

$$\frac{1}{\left(\ln (\ln n)\right)^{\ln n}}=\frac{1}{e^{\ln n\ln (\ln (\ln n))}}=\frac{1}{n^{\ln (\ln (\ln n))}}$$

for n enough big $$\ln (\ln (\ln n))>2$$.and we can say that series converges

Last edited: Nov 18, 2007
4. Nov 19, 2007

### Gib Z

That seems the correct method, though in case you want to clarify it, prove that ln(ln(ln n)) > 2 and is increasing, and also show that you are comparing it to say, zeta(2).

5. Nov 19, 2007

### azatkgz

To prove that it's increasing is it enough?

$$\left(\ln (\ln (\ln n))\right)'=\frac{1}{n\ln (\ln n)}>0$$

6. Nov 19, 2007

### quasar987

you got the derivative wrong. think chain rule.

and in case you're wondering zeta(s)=$\sum$1/n^s, so zeta(2)=$\sum$1/n². for a complete solution, you should specify that you're comparing the general term to that of zeta(2).

Last edited: Nov 19, 2007