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Second rule of comparison in math series

  1. Nov 4, 2004 #1
    I cant understand this [tex]\sum{A_n}\leq \sum{B_n}[/tex] having said this

    if [tex]\sum{B_n}[/tex] converges so does [tex]\sum{A_n}[/tex], okay that

    makes perfect sense but then the second rule of comparison is if [tex]\sum

    {A_n}[/tex] diverges then so does [tex]\sum{B_n}[/tex] diverges too...can

    anyone tell me how that makes sense? A proof maybe..?
  2. jcsd
  3. Nov 4, 2004 #2

    matt grime

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    General result, if P implies Q, then not Q implies not P.

    replace P with sum Bn converges and Q sum An converges.
  4. Nov 5, 2004 #3


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    Given [tex]\sum{A_n}[/tex] does NOT converge.

    Now assume [tex]\sum{B_n}[/tex] DOES converge. Using the theorem you said "makes perfect sense", what does that tell you about [tex]\sum{A_n}[/tex]
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