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Limit comparison test assistance needed please

  1. Jan 2, 2015 #1
    1. The problem statement, all variables and given/known data
    Use the limit comparison test to check for convergence or divergence: Sum from n=1 to infinity of ((2n)^2+5)^-3

    2. Relevant equations
    let lim n to infinity of An/Bn = c
    1) if 0<c<infinity then either both converge or both diverge
    2) if c=0 and sum Bn converges, so does sum An
    3) if c=infinity and sum Bn diverges, so does sum An

    3. The attempt at a solution
    see attached - I'm a little confused. It seems that it would be easy to pick a comparison series to "force" the answer that you want....
     

    Attached Files:

  2. jcsd
  3. Jan 2, 2015 #2

    Stephen Tashi

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    Chosing [itex] b_n = 1 + \frac{1}{\sqrt[3]{n} } [/itex] doesn't let you draw any conclusion because [itex] \sum_{n=1}^\infty b_n [/itex] doesn't converge.
     
  4. Jan 2, 2015 #3
    Doesn't it converge to 1? 1 + 0 = 1? If not, why not? And also what should I choose instead to be bn? Thanks
     
  5. Jan 2, 2015 #4

    Stephen Tashi

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    Don't confuse the convergence of the sequence [itex] \lim_{n \rightarrow \infty} b_n = 1 [/itex] with the convergence of the series [itex] \sum_{n=1}^\infty b_n [/itex]. The part 2 in your statement of the limit comparison tests talks about [itex] b_n [/itex] being summable.

    My guess is [itex] b_n = (\frac{1}{n^2})^{-3} [/itex].

    Edit: Try [itex] b_n = (n^2)^{-3} [/itex]
     
    Last edited: Jan 3, 2015
  6. Jan 3, 2015 #5

    epenguin

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    That looks like a photo taken with i-pad - for presentation the app mentioned below is almost no extra trouble and makes easier on the eye. :oldsmile: DocScan HD

    "But if you do do that, there are apps, at least one I know called DocScan HD which if you are taking these with I-pad or similar cleans them up from the hard-to-read grey-yellow."
     
    Last edited: Jan 3, 2015
  7. Jan 3, 2015 #6

    epenguin

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    If you find it very difficult to prove convergence, hedge your bets and try to test for divergence. I would compare with something like 1/n (which I hope you know is divergent.)
     
  8. Jan 3, 2015 #7

    Mark44

    Staff: Mentor

    Thread closed. Please stop posting messy images of your work. Most of what you posted has been crossed out, and much of the rest is difficult to read.
     
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