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Series convergence

  1. Oct 17, 2007 #1
    It's easy to see that [tex]\sum_{n=2}^{\infty}\frac{1}{lnn}[/tex] does not converge. But what happens to [tex]\sum_{n=2}^{\infty}\frac{1}{(lnn)^k}[/tex] with k > 1 and why?

    Can anybody help?
     
  2. jcsd
  3. Oct 17, 2007 #2
    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.
     
  4. Oct 17, 2007 #3
    Thanks!
     
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