Does the Series 1/[(ln(n))^ln(n)] Converge?

[ryeager]

Homework Statement

Determine whether the series is convergent or divergent.

Sum from n=2 to n = infinity of

1/[(lnx)^lnx]

Homework Equations

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.

 

[sutupidmath]

did you mean here

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

 

[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.

 

[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