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Determine Absolute Convergence, Conditionally Convergent, or Divergent

  1. May 13, 2012 #1
    1. The problem statement, all variables and given/known data
    ##\sum _{n=1}^{\infty }\dfrac {\left( -3\right) ^{n}} {n^{3}}##

    According to Wolfram Alpha the series diverges by the Limit Comparison Test, but I remember that the limit comparison only works with series greater than zero. How is this possible?

    2. Relevant equations



    3. The attempt at a solution
    Also, I have no clue how to use the alternating series test on this one, since it's (-3)^n in the numerator not (-1)^n or (-1)^n+1

    Please, help.
     
  2. jcsd
  3. May 13, 2012 #2

    sharks

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    Isn't that the same as:
    [tex]\sum _{n=1}^{\infty }\frac {3^n(-1)^n}{n^3}[/tex]
     
  4. May 13, 2012 #3
    Yes, sorry about that. Now, I'm trying to figure out what to do with it.
     
  5. May 13, 2012 #4

    sharks

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    Use the Alternating Series Test.
     
  6. May 13, 2012 #5
    Yeah, I'm trying to prove that 3^n > n^3
     
  7. May 13, 2012 #6

    sharks

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    How do you use the Alternating Series Test? There are 2 steps. The first is finding the limit of ##a_n## and testing if it's zero.
     
  8. May 13, 2012 #7
    Step1: lim as "n" approaches infinity of A_n =0
    Step2: A_n > A_(n+1)
     
  9. May 13, 2012 #8

    sharks

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    Work out the first step and verify if you get zero, or not.
     
  10. May 13, 2012 #9
    Yeah, that's what I am trying. Supposedly, I doesn't go to zero. Then from the test of divergence, it diverges. However, I don't know how to test it as n goes to infinity.
    ##\lim _{n\rightarrow \infty }\dfrac {3^{n}} {n^{3}}##

    I'm trying to prove that 3^n is much bigger than n^3; therefore, it diverges.
     
  11. May 13, 2012 #10

    sharks

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    Use L'Hopital's Rule.
     
  12. May 13, 2012 #11
    Great! Thank you.

    ## \lim _{n\rightarrow \infty }\dfrac {3^{n}} {n^{3}}\dfrac {H} {-}\dfrac {3^{n}} {9n}\dfrac {H} {-}\dfrac {n3^{n-1}} {9}##

    Diverges.

    Why would Wolfram Alpha state that by the limit comparison test, the series diverges?
     
  13. May 13, 2012 #12

    sharks

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    Applying L'Hopital's Rule, your numerator is wrong. What is the derivative of ##3^n##? The derivative of your denominator is also wrong.
     
  14. May 13, 2012 #13
    (3^n) ln(3)
     
  15. May 13, 2012 #14

    sharks

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    You'll have to apply L'Hopital's Rule three times to get:
    [tex]\lim_{n\to \infty} \frac{3^n (\ln 3)^3}{6}[/tex]
     
  16. May 13, 2012 #15
    You are correct.
    Why would Wolfram Alpha state that by the limit comparison test, the series diverges?
     
  17. May 13, 2012 #16
    I just put it into WolframAlpha and it said "By the limit test, the series diverges" (not limit comparison).
     
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