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Convergence of a series with tests

  1. May 12, 2012 #1
    Determine whether the series converges or diverges:

    [itex]\sum[/itex] ln k/ k3

    now I said that this series converges by the comparison test, using ln k / k since I know that goes to 0

    Would that be the right logic?
     
  2. jcsd
  3. May 12, 2012 #2

    sharks

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    In the series, k tends towards what value? Your question is missing that information.
     
  4. May 12, 2012 #3
    oh, sorry. k is tending towards ∞
     
  5. May 12, 2012 #4

    LCKurtz

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    You haven't written enough to tell. What comparison test are you using? What convergent series are you comparing it with?
     
  6. May 12, 2012 #5

    I was using a basic comparison test. And I was using ln k/ k as my convergent series. Unless ln k/ k is not a convergent series, but it is a convergent sequence though.
     
  7. May 12, 2012 #6

    LCKurtz

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    Maybe you confuse sequences and series. What makes you think$$
    \sum_{k=1}^\infty \frac {\ln k}{k}$$is a convergent series? Just because the kth term ##\frac {\ln k} k\rightarrow 0## doesn't mean the series converges.
     
  8. May 12, 2012 #7

    Actually I've been trying to straighten out that confusion, but your right I do mix them up sometimes. Ok, well by a comparison test with 1/k, that would mean $$
    \sum_{k=1}^\infty \frac {\ln k}{k}$$ is a divergent series

    but then that doesn't help me here because $$
    \sum_{k=1}^\infty \frac {\ln k}{k}$$

    is larger than
    $$
    \sum_{k=1}^\infty \frac {\ln k}{k^3}$$

    perhaps a limit comparison with

    $$
    \sum_{k=1}^\infty \frac {\ln k}{k}$$ would work?
     
  9. May 12, 2012 #8

    micromass

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    Use that ln(k)≤k for large k.
     
  10. May 12, 2012 #9
    Yes, then that simplifies to 1/k2 which converges based off of p-series.

    Thanks
     
  11. May 12, 2012 #10
    On another note......what's a good way to stop confusing sequences, partial sums, and series?
     
  12. May 12, 2012 #11

    micromass

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    Concentration.
     
  13. May 12, 2012 #12

    sharks

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    :biggrin:
     
  14. May 12, 2012 #13
    I have another quick question about the basic convergence test:


    If lim an as n-->∞ ≠ 0 then Ʃ an will diverge.



    is the an that is being talked about at the beginning of the theorem the "sequence" an or is it the values of the partial sum of all an that they are talking about?
     
  15. May 12, 2012 #14
    The a[itex]_{n}[/itex] you are referring to is a sequence of terms. The k-th partial sum of [itex]\sum[/itex]a[itex]_{n}[/itex] is defined S[itex]_{k}[/itex]=a[itex]_{0}[/itex]+a[itex]_{1}[/itex]+a[itex]_{2}[/itex]+...+a[itex]_{k}[/itex]. I hope this clears up some of the confusion you've been having regarding series, sequences, and partial sums.
     
  16. May 13, 2012 #15

    LCKurtz

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    ##a_n## is the nth term of the series ##\sum_{n=1}^\infty a_n##. Of course if the series starts with ##n=0## it isn't exactly the nth term any more, it's the "general term", but it is still a good way to keep things straight.
     
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