Does \(\sum_{n=2}^{\infty}\frac{1}{(\ln n)^k}\) Converge for \(k > 1\)?

[photis]

It's easy to see that \sum_{n=2}^{\infty}\frac{1}{lnn} does not converge. But what happens to \sum_{n=2}^{\infty}\frac{1}{(lnn)^k} with k > 1 and why?

Can anybody help?

 

[SiddharthM]

i'm pretty sure it diverges for all k>1. this is because n/((lnn)^k) goes to infinity for all integer values of k. THink about this - use L'Hopitals rule and definition of a limit for a formal proof using the comparison test with the harmonic series.

 

[photis]

Thanks!