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Homework Help: Convergent or Divegent Series?

  1. Dec 9, 2008 #1
    1. The problem statement, all variables and given/known data
    [tex]\sum{(ln(k))/(\sqrt{k+2})}[/tex], with k starting at 1 and going to [tex]\infty[/tex]

    2. Relevant equations
    Does this series converge or diverge? Be sure to explain what tests were used and why they are applicable.

    3. The attempt at a solution
    Okay, my TA got that this diverges, but I got that it converges by simply taking the limit as k goes to [tex]\infty[/tex] and applying L'Hopital's rule. I also plugged the function into my calculator and it seems to converge at y=0, which is corroborates what I got with L'Hopitals rule. What did you guys get? Any help is greatly appreciated.
  2. jcsd
  3. Dec 9, 2008 #2


    User Avatar
    Staff Emeritus
    Science Advisor
    Gold Member

    It looks like you proved the individual terms go to 0, which isn't what you're trying to do. The definition of a series is you take the limit of the partial sums. And where does y come into the series?

    Try comparing the series to [tex] \frac{1}{\sqrt{k+2}}[/tex] whose convergence/divergence is easier to find
  4. Dec 9, 2008 #3
    Right, I forgot about the partial sums. Thanks, that helped a lot. But isn't [tex]1/(\sqrt{k+2})[/tex] smaller than [tex](ln(k))/(\sqrt{k+2})[/tex]?
    Last edited: Dec 9, 2008
  5. Dec 9, 2008 #4


    Staff: Mentor

    Yeah, it is. You've been handed a clue for free. If you can say something about what [tex]\sum 1/(\sqrt{k+2})[/tex] does, then maybe you will know something about the series you're really interested in.
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