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Sequence convergence?

  1. Sep 24, 2009 #1
    1. The problem statement, all variables and given/known data
    If {s_n} is a bounded sequence of real numbers and {t_n} converges to 0, prove that {s_n*t_n} converges to 0.


    2. Relevant equations


    3. The attempt at a solution
    I know that since s_n is bounded, ||s_n|| <= M, but I cannot seem to find a way to make this work into the proof for convergence. Any help will be appreciated.
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 24, 2009 #2

    LCKurtz

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    Since you have |sn| ≤ M what can you say about |sntn|?
     
  4. Sep 24, 2009 #3
    You can conclude that ||s_n*t_n||<=||t_n|*M. I know that t_n < epsilon, but I am having trouble working around to that.
     
  5. Sep 24, 2009 #4

    LCKurtz

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    OK. Can you use that and the fact you know |tn| → 0, to claim if you were given ε > 0, you can make

    |tnsn| < ε ?
     
  6. Sep 24, 2009 #5
    Thanks; you have helped immensely.
     
  7. Sep 24, 2009 #6
    Sorry, but I must still be stumped. I wrote the following:
    Proof: Let \epsilon > 0. By the definition of bounded, |s_n|<=M, where M is a real number and n is an integer. This implies that |s_n*t_n|<=|t_n|*M. Since we know that t_n approaches 0 and \epsilon > 0, it follows that |s_n*t_n| < \epsilon. We can therefore conclude that s_n*t_n approaches 0. End of proof.

    I can't seem to get the "we need to find a positive integer N such that |s_n*t_n|<\epsilon" part. I can't find N. I seem to be caught up in the abstractness of it. I will certainly appreciate just a tiny bit more help. Thank you.
     
  8. Sep 24, 2009 #7
    I think you have most of it. First of all, if [tex]\lim t_n = 0[/tex], then what can you say about it? So, let [tex]\forall \varepsilon > 0[/tex]. Since [tex](s_n)[/tex] is bounded, then [tex]\exists M > 0[/tex] such that [tex]|s_n| \leq M[/tex] for [tex]\forall n \in \mathbb{N}[/tex]. Since [tex]\lim t_n = 0[/tex], [tex]\exists N[/tex] such that [tex]\forall n > N[/tex], we have [tex]|t_n - 0| < \varepsilon /M [/tex]. Note that we have [tex]\varepsilon / M > 0[/tex] and this is perfectly valid (and the key to why your proof is a little lacking --- can you think of why we can write [tex]\varepsilon / M[/tex] instead of just the usual [tex]\varepsilon[/tex] on the right hand side of the above?).

    Then it follows that
    [tex]|s_n t_n - 0| \leq M|t_n| < M \cdot \frac{\varepsilon}{M} = \varepsilon[/tex], for [tex]\forall n > N[/tex]. This completes the proof. :)
     
  9. Sep 24, 2009 #8
    Ah, yes, it does!!! Thank you so much. This time I really do have it! Have a nice evening and thank you.
     
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