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Homework Help: Properly Divergent Sequences

  1. Oct 29, 2008 #1
    1. The problem statement, all variables and given/known data
    Suppose that (x_n) is a properly divergent sequence, and suppose that (x_n) is unbounded above. Suppose that there exists a sequence (y_n) such that limit (x_n * y_n) exists. Prove that (y_n) ===> 0.


    2. Relevant equations
    (x_n) ===> 0 <====> (1/x_n) ===> 0


    3. The attempt at a solution
    One can say with certanty that (y_n) must be bounded, as if it weren't, for all K in Naturals, there exists a b_1 in (x_n) > |K| and b_2 > |K|, and there product is unbounded.

    If (y_n) is bounded, and does not converge to 0, then... what?

    That's where I'm stuck. How do I finish this?

    Thanks.
     
  2. jcsd
  3. Oct 29, 2008 #2

    Dick

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    If y_n does not converge to zero then there is an e>0 such that for all N there is an n>N such that |y_n|>e. If x_n is unbounded, what does this tell about y_n*x_n?
     
  4. Oct 30, 2008 #3
    When I was trying to prove it directly (for fun), I met some problems.
    What is a *properly* divergent sequence?( I cannot find its definition in books)
    If x_n is defined as follows:
    x_n = 0 when n is even, x_n = n when n is odd
    is it of such kind?
    If so, define y_n as:
    y_n = 1 when n is even, y_n = 0 when n is odd

    does this gives x_n*y_n = 0, as a counter??
     
    Last edited: Oct 30, 2008
  5. Oct 30, 2008 #4

    HallsofIvy

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    The crucial part of the problem is "suppose that (x_n) is unbounded above". Your example does not satisfy that.
     
  6. Oct 30, 2008 #5

    Office_Shredder

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    A properly divergent series is one such that

    (mathematics) A series whose partial sums become either arbitrarily large or arbitrarily small (algebraically).

    So turning each partial sum into an element of the sequence, I believe a properly divergent sequence is one in which for all M there exists k s.t. j>k => |xj|>M
     
  7. Oct 30, 2008 #6
    Thanks..I forgot to search the web.... That would make sense. So a direct proof is also not hard.

    BTW, to HallsofIvy, my x_n do satisfy the unboundedness, IMO.
     
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