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Convergence/Divergence of series; conflicting answers

  1. Nov 13, 2012 #1
    1. The problem statement, all variables and given/known data
    Simple question. Find the convergence or divergence of this term.

    3. The attempt at a solution
    I tried using the comparison test with 1/sqrtn, which is a divergent p series (p=1/2), and then with 1/2^n, which is a convergent geometric series. I don't know why the answers should be conflicting, since it is clear both of my comparison series are bigger than the original.

    Can someone help me? I'm probably doing something very stupid.

    http://i47.tinypic.com/jj19xu.jpg
     
    Last edited by a moderator: Nov 13, 2012
  2. jcsd
  3. Nov 13, 2012 #2

    SammyS

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    Assuming that the terms of the series are positive ...

    When using the comparison test to check for divergence, you want your series to be larger term by term than the known divergent series.
     
  4. Nov 13, 2012 #3

    HallsofIvy

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    In order to prove that a positive series [tex]\{a_n\}[/tex] diverges, by the comparison test, you must show there is a divergent series [tex]\{b_n\}[/tex] such that [tex]a_n\ge b_n[/tex]. In order to show that a positive series [tex]\{a_n\}[/tex] converges, by the comparison test, you must show that there is a convergent sequence [tex]\{b_n\}[/tex] such that [tex]a_n\le b_n[/tex].
     
  5. Nov 14, 2012 #4
    That is precisely the problem. Both of my comparisons are bigger than the original series, but one is divergent and the other is convergent.

    Can I say that just because one of my comparisons converges, that my original series also converges (and that the divergent series comparison provides no information?).
     
  6. Nov 14, 2012 #5

    Dick

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    That IS precisely the problem. Your comparison 1/sqrt(n) is bigger than the original series. You take any series whatsoever and it's easy to find a divergent series that's bigger. So yes, no information there.
     
  7. Nov 14, 2012 #6

    SammyS

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    So, find a known divergent series that's than your series, term by term.

    Otherwise, use a different method to show that your series is divergent.
     
  8. Nov 14, 2012 #7
    Hmm, I am sorry to say, but I am still kind of confused.

    Now that I am rethinking the problem, I see that perhaps only one of them suggests convergence. If the bigger term converges, then the smaller term will converge too.

    1/sqrtn is bigger, but it is divergent, so tells me nothing about my series.
    1/2^n is also bigger, but because it is convergent, I know that my series is convergent too.

    So I think one way to look at it is that it doesn't matter if 1/sqrt diverges, because that information is not useful. All I need to do is find a bigger series that converges, or a smaller one that diverges. A bigger one that diverges doesn't help me. Please correct me if I'm wrong.

    In terms of using a comparison that is "bigger term by term," does this mean that I should not have tried the comparison with 1/sqrtn and simply went to 1/2^n?
     
  9. Nov 15, 2012 #8

    SammyS

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    I misunderstood the problem you're working on. (Must not have looked at the link!)

    Do you know if [itex]\displaystyle \ \ \sum_{n=1}^{\infty} \frac{1}{\sqrt{n}+2^n}\ \ [/itex] converges ?

    If it does, the find a known convergent series which is larger, term by term, than your series.

    Note: [itex]\displaystyle \ \ \sum_{n=1}^{\infty} \frac{1}{2^n}\ \ [/itex] is not larger term by term.


    If it diverges, the find a known divergent series which is smaller, term by term, than your series.
     
  10. Nov 15, 2012 #9

    Dick

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    Nothing you are saying here sounds confused.
     
  11. Nov 15, 2012 #10

    Curious3141

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    Isn't it? I thought it was.:confused:
     
  12. Nov 15, 2012 #11

    Curious3141

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    To prove convergence for a series of all positive terms, you need to find another *convergent* series of positive terms for which every term is *larger* than the original series. Think about this as finding an upper bound for your original series. If the upper bound is finite, your original series has to converge.

    Conversely, to prove divergence, you need to find a *divergent* series which is *smaller*, term by term than the original series. Think about this as finding a lower bound for your original series. If the lower bound is infinite, your original series has to diverge.

    (In both cases, if you cannot find a corresponding series where every term meets the condition, it suffices to find one where every term from a certain term onward meets the condition. But this is small print, and not needed here).

    So the first test you did doesn't add any value. You compared against a divergent series *but* your original series was smaller, term by term than this divergent series. You cannot conclude anything from this.

    The second test has value. You compared against a convergent series, *and* your original series is smaller, term by term than this convergent series. You can draw a conclusion from this.

    There is no contradiction here. It's just that one comparison was useful, the other not.
     
  13. Nov 15, 2012 #12

    Ray Vickson

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    The series [tex]\sum_{n=1}^{\infty} \frac{1}{2^n}[/tex] IS larger, term-by-term, than the original series, because [tex] \sqrt{n} + 2^n > 2^n[/tex] for all n ≥ 1.

    RGV
     
  14. Nov 15, 2012 #13

    SammyS

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    DUH!

    Of course it is larger term by term!

    What was I thinking?
     
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