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Series root or ratio test?

  1. Mar 23, 2010 #1
    1. The problem statement, all variables and given/known data

    Determine whether the series converges or diverges.

    [tex]\sum (\sqrt {k} - \sqrt {k - 1})^k[/tex]


    2. Relevant equations



    3. The attempt at a solution

    [tex](a_k)^\frac{1}{k} = \sqrt{k} - \sqrt{k - 1} [/tex]

    What do I do here..?

    [tex]= \frac{1}{\sqrt{k} + \sqrt{k-1}} \to 0 ?[/tex]
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Mar 23, 2010 #2

    Mark44

    Staff: Mentor

    I would start with this:
    [tex]\sqrt{k} - \sqrt{k - 1} = \frac{(\sqrt{k} - \sqrt{k - 1})(\sqrt{k} + \sqrt{k - 1})}{\sqrt{k} + \sqrt{k - 1}}[/tex]
     
  4. Mar 23, 2010 #3
    So that is
    [tex]
    \sqrt{k} - \sqrt{k - 1} = \frac{(\sqrt{k} - \sqrt{k - 1})(\sqrt{k} + \sqrt{k - 1})}{\sqrt{k} + \sqrt{k - 1}}
    = \frac {\sqrt{k}^2 - \sqrt{k-1}^2}{\sqrt{k} + \sqrt{k - 1}} = \frac{1}{\sqrt{k} + \sqrt{k-1}} [/tex]
     
  5. Mar 23, 2010 #4
    Then

    [tex]\frac{1}{\sqrt{k} + \sqrt{k-1}} < \frac{1}{\sqrt{k}} = ( \frac{1}{k})^{\frac{1}{2} [/tex]? then converges?
     
  6. Mar 23, 2010 #5

    Mark44

    Staff: Mentor

    You are apparently confusing yourself. You started with the root test, not the comparison test.

    BTW, the series whose general term is 1/sqrt(k) diverges, but that's not relevant to what you're doing.
     
  7. Mar 23, 2010 #6
    Ok so I go here [tex]\frac{1}{\sqrt{k} + \sqrt{k-1}}[/tex] and then I'm kinda stuck
     
  8. Mar 23, 2010 #7

    Mark44

    Staff: Mentor

    Take the limit as k --> infinity. What do you get? Why is this limit important to you? What did you start out doing in your first post?
     
  9. Mar 23, 2010 #8
    I get 0? So the ratio is < 1 so its converges?
     
  10. Mar 23, 2010 #9

    Mark44

    Staff: Mentor

    The limit is < 1. Try not to confuse yourself into think you are working with the ratio test - here it's the root test, so the fact that you are finding the limit of a fraction is not relevant.
     
  11. Mar 24, 2010 #10
    Ok so since [tex]a_k < \mu^k[/tex] and [tex] \mu^k[/tex] converges [tex] a_k [/tex]converges
     
  12. Mar 24, 2010 #11

    Mark44

    Staff: Mentor

    I don't know -- what's [itex]\mu[/itex]? That's the first time it has appeared in this thread.

    zeion, you need to step back and take a bigger-picture view of what you're doing. You seem to be getting lost in minute details, and losing track of the purpose of the details.

    Maybe it will help you grasp the overall strategy by answering these questions.
    1. What test are you using?
    2. When you use this test, what result indicates that the series you're testing converges?
      What result indicates that the series diverges?
    3. What result did you get?
    4. What can you conclude about the series you are testing?
     
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