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Homework Help: Infinite Series & Limits

  1. May 8, 2007 #1
    1) Determine whether the infinite series

    Sigma (k^2-1) / (3k^4 + 1)
    converges or diverges.

    [My immediate thought was to use the "limit comparsion test", but this test requires all terms to be positive. However, the first term (put k=0) is definitely negative...what should I do? Can I still use the limit comparsion test, and if not, what other tests can I use?]

    2) Evaluate
    lim [t^2 - t^3 sin(1/t)]

    [When I direct substitute, I get ∞-∞*0, and I have no clue how to solve this problem...any hints?]

    Thank you for your help!:smile:
  2. jcsd
  3. May 8, 2007 #2
    let [tex]a_{n} = \frac{k^2 - 1}{3k^4 + 1}[/tex]
    and [tex]b_{n} = \frac{1}{k^2}[/tex]
    then show that [tex]a_{n} > b_{n}[/tex] for every n

    Okay, I finally got it...
    Last edited: May 8, 2007
  4. May 8, 2007 #3
    But if an>bn and bn converges, it tells you nothing whether an converges or not...how does that help?
    And also, it doesn't look very straight foward to me to prove that an>bn
  5. May 8, 2007 #4
    1) Also, I am still wondering whether I can apply "limit comparsion test" to this problem or not...does anyone know?

  6. May 8, 2007 #5


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    Staff Emeritus
    Science Advisor
    Gold Member

    This is where convergence is beautiful. It doesn't matter if the first term is negative; just look at the sum

    term one + series starting from term two.

    Clearly this sum exists iff your original series converges (if that's not obvious, prove it)

    BTW, I think [tex]a_n<b_n[/tex] as posted above, because [tex]a_n[/tex] approximates [tex]\frac{1}{3k^2}[/tex]

    (in fact, it's always less than that.... you should be able to show that too)
  7. May 9, 2007 #6
    So when I am trying to see whether an infinite series converges or diverges, I can always change the lower index of summation whenever I want?

    an approximates 1/3k^2, but how can I know which one is larger?

  8. May 9, 2007 #7


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    Science Advisor
    Homework Helper
    Gold Member

    That's right; convergence or divergence of a series is independent or the starting index.
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