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Convergence Tests the Comparison Test

  1. Jan 19, 2009 #1
    hey guys i'm kind of stuck on this idea of the comparison test. i understand the process and how it is done but i don't seem to understand how to find the second series(typically called bn) to compare with the original. get what i mean?

    thanks! =)
     
  2. jcsd
  3. Jan 19, 2009 #2
    Well the comparison test says that if you have [itex]0\leq a_k \leq b_k[/itex], then if

    [tex]\sum_{k=1}^{\infty}b_k<\infty \quad \Rightarrow \quad \sum_{k=1}^{\infty}a_k <\infty[/tex]

    and if

    [tex] \sum_{k=1}^{\infty}a_k=\infty \quad \Rightarrow \quad \sum_{k=1}^{\infty}b_k=\infty[/tex]

    So you need to find either a smaller [itex]a_k[/itex] or the larger [itex]b_k[/itex] in either case, but as you may see there isn't really any prescribed method of finding it. You just need to look at the series and think of what you could compare it to that is smaller or larger depending on the case. You can see why it's useful to know the convergence/divergence properties of many common infinite series.
     
  4. Jan 19, 2009 #3
    so you can technically compare it to anything?
     
  5. Jan 19, 2009 #4
    Yes, if after large [itex]k[/itex] the series fits the necessary condition of being larger or smaller. Of course this cannot be used for some obvious series which wont work, like alternating, which have their own tests.
     
  6. Jan 19, 2009 #5
    just curious, why is the alternating series test different?
     
  7. Jan 19, 2009 #6
    Well think of the alternating series, something like

    [tex] \sum_{k=1}^{\infty} (-1)^n a_k [/tex]

    with [itex]a_k[/itex] always positive. This will jump back and forth from negative to positive on the number line as it progresses so most of the time a comparison test with these wont work (you would need to find something smaller or larger but you can see why this would be difficult).
     
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