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Homework Help: Series question

  1. Apr 21, 2008 #1

    daniel_i_l

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    1. The problem statement, all variables and given/known data
    Lets say that I have some sequence [tex](a_n)[/tex] which converges to 0 at infinity and that for all n [tex]a_{n+1} < a_n [/tex] but the sequence [tex](a_n)[/tex] diverges. Now I know that the series
    [tex](cos(n) a_n) [/tex] converges but can I use the following argument to prove that
    [tex]|cos(n) a_n| [/tex] doesn't converge:
    [tex] |cos(n) a_n| >= {cos}^{2}(n) a_n = {a_n}/2 + {(cos(2n)) a_n}/2 [/tex]
    And since [tex]{(cos(2n)) a_n}/2 [/tex] converges and [tex] {a_n}/2 [/tex] diverges
    [tex] {cos}^{2}(n) a_n [/tex] diverges and so [tex] |cos(n) a_n| [/tex] diverges.
    Is that always true?
    Thanks.
     
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  3. Apr 21, 2008 #2

    Dick

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    That seems ok, provided you can prove cos(2n)*a_n converges. Are you using Abel-Dedekind-Dirichlet (summation by parts)? I just figured out how it works, so I had to ask.
     
  4. Apr 22, 2008 #3

    daniel_i_l

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  5. Apr 22, 2008 #4

    Dick

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  6. Apr 22, 2008 #5

    daniel_i_l

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    Well I know that cos(n) has bounded partial sums and cos(2n) = 2(cos(n))^2 - 1 so that means that cos(2n) also has bounded partial sums right?
     
  7. Apr 22, 2008 #6

    Dick

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    Not at all! cos(n)^2 doesn't have bounded partial sums and neither does 1! How did you show cos(n) had bounded partial sums?? Try and apply the same technique to cos(2n).
     
  8. Apr 22, 2008 #7

    HallsofIvy

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    You mean series rather than that last "sequence", right?
     
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