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Homework Help: In search of a counter-example

  1. Feb 7, 2009 #1
    1. The problem statement, all variables and given/known data

    Is it true that if {pn}{qn}-> 0, that either pn or qn converges to 0?

    2. Relevant equations

    3. The attempt at a solution So far I've made an argument for why it doesn't have to follow (Because we don't know that both sequences converge, and we could have two divergent sequences or a divergent sequence and a convergent sequence whose product converges, for example the alternating harmonic series), However I can't think of any particular counter-example where the product converges to 0.
  2. jcsd
  3. Feb 7, 2009 #2
    Try to arrange it so that the even terms of {p_n} converge to zero, but {p_n} does not, and conversely, the odd terms of {q_n} converge to zero, but {q_n} does not.
  4. Feb 7, 2009 #3
    That can't happen though can it? If {p_n}(even)-> 0, then the even terms form a subsequence, so wouldn't the sequence have to converge to 0 also?
  5. Feb 7, 2009 #4
    nevermind, I got it I think.


    My concern isn't relevant because they diverge.
  6. Feb 7, 2009 #5


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    Yes, if {x_n} converges to x and {y_n} converges to y, then {x_ny_n} must converge to xy. So if {x_n} and {y_n} converge AND {x_ny_n} converges to 0, either x or y must be 0.

    But you can find sequences that do NOT converge but {x_ny_n} converges to 0.
  7. Feb 7, 2009 #6
    Hm... what about if {x_ny_n} converges, but only {x_n} or {y_n} converges?

    Would that be possible, or would {x_n}-> x imply {y_n} also converged as well?
  8. Feb 7, 2009 #7


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    n*(1/n) converges, right?
  9. Feb 8, 2009 #8
    As dick showed, it is possible to have {x_n*y_n} and {x_n} converge, but {y_n} diverge. However, this is only possible if [itex]x_n \rightarrow 0[/itex]. Because if [itex]x_n y_n \rightarrow L[/itex] and [itex]x_n \rightarrow x \neq 0[/itex], then applying the main limit theorem shows that:

    [tex]\lim y_n = \lim \frac{x_n y_n}{x_n} = \frac{\lim x_n y_n}{\lim x_n} = \frac{L}{x}[/tex]

    i.e. [itex]y_n \rightarrow \frac{L}{x}[/itex]
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