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Showing Convergence

  1. Dec 9, 2007 #1
    Show that [tex]\sum_{k=1}^{\infty}\frac{k+4}{k^3}[/tex] converges.

    I thought I would try the limit comparison test by using [tex]a_k=\frac{k+4}{k^3}[/tex] and [tex]b_k=\frac{1}{k^3}[/tex].

    I thought a_k looked similar to power series so that if the lim ask-->infinity of a_k/b_k is finite and > 0 then since the p-series (for p>1) converges, than so must a_k,

    But I am getting an infinite limit.

    So my questions are:

    1.) Where is my reasoning flawed?

    2.) What is the correct approach?

    ~Casey
     
  2. jcsd
  3. Dec 9, 2007 #2

    cristo

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    Your original series can be written [tex]\sum_{k=1}^\infty\frac{1}{k^2}+\frac{4}{k^3}[/tex]. Does this give you a hint as how to proceed?
     
  4. Dec 9, 2007 #3

    mjsd

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    use [tex]b_k = 1/k^2[/tex] instead
     
  5. Dec 9, 2007 #4
    I like this, but we shouldn't need to use partials.

    I see that this will give me a limit of 1...but you have not given me any justification as to WHY I should or even can use p=2 as opposed to p=3 when the original series had p=3 ?

    Someone please elaborate.

    Casey
     
  6. Dec 9, 2007 #5

    mjsd

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    Read the limit comparsion test description again....where does it say that the p-series used must be related to the original series in any way?


    EDIT: all you require is that both a_k and b_k are positive for all k
     
  7. Dec 9, 2007 #6
    Well, my instructor is forever implying that we should use a series that relates. But you are correct, the description does not say that it needs to.

    That being said, why can I not use p= anything, i.e, 3 in this case? I s can use anything...why NOT 3?

    Thanks for your patience,
    Casey
     
  8. Dec 9, 2007 #7

    mjsd

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    you aim here is to find a series b_k that will help you show whether sum of a_k is convergent or not. Note that a_k ~ 1/k^2 when k is large, that's why using b_k = 1/k^2 will help because we know that a_k/b_k ~ 1/k^2 / (1/k^2) ~ finite. or at least the first thing one would try

    yeah.. in a sense, that's why you should use a series that relates.... but it is more subtle than just: "since that you have a k^3 on the bottom for a_k so it would means p=3..."
     
  9. Dec 9, 2007 #8
    I am still not clear on why if all we need to do is use a b_k such that we know the outcome of it....and we know [tex]\sum b_k=\sum_1^{\infty}\frac{1}{k^3}[/tex] converges and we also know that
    [tex]\sum a_k=\sum_1^{\infty}\frac{k+4}{k^3}[/tex] converges...then why does there the limit at infty of a_k/b_k not yield a finite number...

    I mean I know that it doesn't because the limit is what it is....but shouldn't my selection of b_k, by the definition of the limit comparison test, be completely arbitrary?

    Thanks again,
    Casey
     
    Last edited: Dec 9, 2007
  10. Dec 9, 2007 #9

    mjsd

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  11. Dec 9, 2007 #10

    cristo

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    The point is that you're looking at the test the wrong way around. The limit comparison test says that if [itex]\lim_{k\to\infty}\frac{a_k}{b_k}>0[/itex] , then a_k converges iff b_k converges. If the limit is infinite, and b_k converges, it does not tell you anything about a_k.
     
    Last edited: Dec 9, 2007
  12. Dec 9, 2007 #11
    So if I know the outcome of b_k but the lim of a_k/b_k is infinite...then I need to make another comparision since nothing can be learned from this.

    Does that sound correct?

    Casey
     
  13. Dec 9, 2007 #12

    cristo

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    If the limit is infinite and b_k converges, then you need another test. If the limit is infinite and b_k diverges, then as stated on mjsd's link above, a_k also diverges.
     
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