# Homework Help: Series 5.

1. Nov 18, 2007

### azatkgz

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

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

3. The attempt at a solution

for $$u=\ln (\ln n)$$

$$\sum_{u=\ln (\ln 3)}^{\infty}\frac{1}{e^{u^2}}$$

from Root Test

$$\lim_{u\rightarrow\infty}\sqrt{\frac{1}{e^{u^2}}}=\lim_{u\rightarrow\infty}\frac{1}{e^u}=0<1$$
series converges

2. Nov 18, 2007

### quasar987

That u substitution does not give back the original series. ln(ln(3))+1 does noe equal ln(ln(4)).

Try simply the comparison criterion. Hint: (logx)²/x -->0, so (logx)²<x for x large enough.

3. Nov 18, 2007

### azatkgz

$$\ln^2x<x$$

for x=lnn

$$\ln^2(\ln n)<\ln n$$

$$e^{\ln^2(\ln n)}<e^{\ln n}=n$$

$$e^{\ln^2(\ln n)}<n\rightarrow \frac{1}{e^{\ln^2(\ln n)}}>\frac{1}{n}$$

diverges

4. Nov 18, 2007

### quasar987

good!

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