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Use the Limit Comparison Test to determine the series' convergence?

  1. Nov 10, 2013 #1
    1. The problem statement, all variables and given/known data

    Use the Limit Comparison Test to determine if the series converges or diverges:
    Ʃ (4/(7+4n(ln^2(n))) from n=1 to ∞.

    (The denominator, for clarity, in words is: seven plus 4n times the natural log squared of n.)

    2. Relevant equations

    Limit Comparison Test:
    Let Σa(n) be the original series, and Σb(n) be the comparison series.
    1. If the lim(n->∞) a(n)/b(n) is a positive number, but not ∞, then a(n)'s convergence/divergence is the same as b(n)'s.
    2. If the lim(n->∞) a(n)/b(n) = 0, then if b(n) is convergent, so is a(n), but if b(n) is divergent, the result is inconclusive.
    3. If the lim(n->∞) a(n)/b(n) = ∞, then if b(n) is divergent, so is a(n), but if b(n) is convergent, the result is inconclusive.

    3. The attempt at a solution

    I need to use a p-series to determine if this series diverges or converges, but every p-series I have tried gives the result that makes it inconclusive.

    Here is the issue:
    When I make b(n) = 1/n (p-series that diverges), I get a limit of 0, which is inconclusive (via case 2).
    When I make b(n) = 1/(n^2) (p-series that converges), I get a limit of ∞, which is inconclusive (via case 3).

    No matter what p-series I try, I get an inconclusive case...

    Can anyone find a comparison p-series that (using the Limit Comparison Test) works?? My instructor said a p-series will work, but everything I try does not.

    Please show all steps. Thank you!
     
  2. jcsd
  3. Nov 10, 2013 #2

    LCKurtz

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    I don't know about the p series thing, but you might try a limit comparison test with ##b_n=\frac 1 {n\ln^2 n}## which series you can try the integral test on. You may have to start at ##n=2## for the second series and integral test.
     
  4. Nov 11, 2013 #3
    I know what you are saying, but my instructor is expecting a p-series, and I have to use the Limit Comparison Test. I really don't know why my instructor says that there is one, but that has to mean there is, unless there's a mistake.

    What do I do if no others reply? Can I re-post this thread?
     
  5. Nov 11, 2013 #4
    My instructor must be mistaken, because everyone that replies to my forum threads on this question says to try what you said. I'll just do that and explain it to my instructor next class.

    Thank you
     
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