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Comparison Test Problem

  1. Nov 8, 2011 #1
    1. The problem statement, all variables and given/known data

    Determine whether the series converges or diverges.

    What I would like is some type of information on how to continue the problem.

    2. Relevant equations


    Ʃ √(n+1)/2n^2+n+1
    n=1

    3. The attempt at a solution

    I was thinking of doing a comparison test by doing

    √(n+1)/2n^2+n+1 < √(n+1)/2n^2 or √(n+1)/2n^2+n+1 < 2n(n+1)

    After that, I end up being stumped by the problem.
     
  2. jcsd
  3. Nov 9, 2011 #2
    Remember what the Comparison Theorem states:

    Suppose that we have two series Ʃa and Ʃb with an,bn≥0 for all n and bn≥ an for all n. Then,

    If Ʃb is convergent then so is Ʃa .
    If Ʃa is divergent then so is Ʃb .

    So let 1/2n2 be Ʃb. Since Ʃb is greater than the given series and we know it converges because of p-series, what can we assume about the original series?
     
  4. Nov 9, 2011 #3

    Mark44

    Staff: Mentor

    You need more parentheses. Without them, the expression in your sum does not mean what you think. A literal reading would give this:
    [tex]\frac{\sqrt{n+1}}{2}\cdot n^2 + n + 1[/tex]

    Since your intent was that 2n2 + n + 1 was in the denominator, write the expression being summed like this:
    √(n+1)/(2n^2+n+1)

    How did you get this one? √(n+1)/2n^2+n+1 < 2n(n+1)
    You're on the right track with the first inequality.
     
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