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Infinite Series. Some confusion with terminology

  1. Dec 2, 2007 #1
    So I am supposed to show that the Infinite series [tex]\sum^{\infty}_{k=1}\frac{3}{k+4}[/tex] does not converge using any method.

    Now, my question: Is [tex]\frac{3}{k+4}[/tex] the General term?

    I will wait for a response before I continue, for it may eliminate another question regarding the General Term and Closed Form....

    Thanks,
    Casey
     
  2. jcsd
  3. Dec 2, 2007 #2

    Dick

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    Yes, I'd call that the general term. As for a test, you could try an integral test. More simply a comparison test, if you happen to know 1/k diverges.
     
  4. Dec 2, 2007 #3
    Right. I did the integral test. But I wanted to do the "By comparison" test as it is definitely looks simpler.

    I have not used the comparison method though. Do I have to show that a "Larger series" diverges and so then the "smaller" series diverges?

    Thanks,
    Casey
     
  5. Dec 2, 2007 #4

    Dick

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    Umm. If the "smaller" diverges, then the "larger" series diverges. Is that what you meant to say?
     
  6. Dec 2, 2007 #5
    Sure why not! That makes me feel better since 3/(k+4) >1/k for k>2.

    And since removing a finite amount of terms does not affect (effect?) converence/divergence then since 1/k diverges, 3/(k+4) must also diverge. Sound good?

    Also. If you have time, what is Closed Form? And when, if ever, would I apply it? I know that it is rare that you can find a closed form, but i am a little confused as to what it is.

    I am looking for an example from the text now to help ....

    Thank you,
    Casey
     
  7. Dec 2, 2007 #6
    continued from above...

    Here's is one:

    "Find a closed form for the nth partial sum and determine if the series converges/diverges by calculating the limit of the nth partial sum."

    [tex]2+\frac{2}{5}+\frac{2}{5^2}+...+\frac{2}{5^{k-1}}+...[/tex]

    Now.....what is [tex]\frac{2}{5^{k-1}}[/tex] is that not the General term?
     
  8. Dec 2, 2007 #7

    Dick

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    Sounds good. That is the 'general term'. But you can write an explicit formula for the sum of the first n terms. It's a geometric series. If you can write a nice formula, that's called 'closed form'.
     
  9. Dec 2, 2007 #8
    And why do we want the closed form? Couldn't I just use the integral test or compare this one? What does a formula do for me that these methods cannot (out of curiousity)?
     
  10. Dec 2, 2007 #9

    Dick

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    The integral test or any other test will just tell you that a limit exists. It won't tell you what the limit is. A closed form can tell you what the limit is.
     
  11. Dec 3, 2007 #10
    Ah Yes...that is right.

    Thanks!
    Casey
     
  12. Dec 3, 2007 #11
    The 'general term' is essentially the 'k-th' term (if you are using the variable k). That is, you can use the 'general term' and obtain any specific term simply by substituting in the appropriate value of k.

    Regarding 'closed form': not quite. The 'closed form' of a series is a nice formula for the ENTIRE sum, not just the sum of the first n terms (that's the 'n-th partial sum'). There is a difference, in particular if your sum is infinite. For example,

    [tex]\frac{1}{1-x} = \sum_{k=0}^\infty x^k \text{ provided } |x|<1[/tex]

    The formula [tex]\frac{1}{1-x}[/tex] is the closed form of the series on the right hand side.
     
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