A tough convergence problem

1. Oct 29, 2008

ryeager

1. The problem statement, all variables and given/known data
Determine whether the series is convergent or divergent.

Sum from n=2 to n = infinity of

1/[(lnx)^lnx]

2. Relevant equations

3. The attempt at a solution

So far I've tried the comparison test, but all I can reach is that the series is less than the harmonic series, which of course doesn't help. Any help would be appreciated greatly, thanks.

2. Oct 29, 2008

sutupidmath

did you mean here

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

3. Oct 29, 2008

Dick

Let's try comparing it with 1/n^2. Remember, you can discard any finite number of terms when you do a comparison test. Try to figure out the limit n->infinity n^2/(log(n)^log(n)). Hint: take the log of the ratio.

4. Oct 30, 2008

boombaby

There's a theorem said,
if a1>=a2>=.....>=0, then $$\sum^{\infty}_{n=1}a_{n}$$ converges if and only if $$\sum^{\infty}_{k=0} 2^{k}a_{2^{k}}$$ converges.
This will make the original series to a better-looking one