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Sequence proof

  1. Sep 2, 2011 #1
    If (b[itex]_{n}[/itex]) is a bounded sequence ad lim(a[itex]_{n}[/itex])=0 show that lim(a[itex]_{n}[/itex]b[itex]_{b}[/itex]) =0

    Pf/

    Let b[itex]_{n}[/itex] be bounded and the lim(a[itex]_{n}[/itex])=0. Since b[itex]_{n}[/itex] is bounded we know that [itex]\exists[/itex] a real number M [itex]\ni[/itex] |b[itex]_{n}[/itex]|<M for all n[itex]\in[/itex][itex]N[/itex] and we also know that |a[itex]_{n}[/itex]|< [itex]\epsilon[/itex] for all [itex]\epsilon[/itex]>0.


    My problem is how do I go from here. I don't believe you can say that the lim(b)*lim(a)=lim(ab)=0 because we don't know what the lim(b) is
     
  2. jcsd
  3. Sep 2, 2011 #2
    Here's how to start: Let [itex]\epsilon > 0[/itex]. Then we have to show that there exists an N > 0 such that if n > N, then [itex]|a_nb_n| < \epsilon[/itex].

    Observe that [itex]|a_nb_n| = |a_n| |b_n|[/itex]. Also, [itex]|b_n| < M[/itex] for all n and we can "make [itex]|a_n|[/itex] as small as we like" for sufficiently large n. Can you put the pieces together from here?
     
  4. Sep 2, 2011 #3
    do you just say for any b[itex]_{n}[/itex] choose an a[itex]_{n}[/itex] such that b[itex]_{n}[/itex] * a[itex]_{n}[/itex] <[itex]\epsilon[/itex]
     
  5. Sep 2, 2011 #4

    vela

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    The bolded part isn't correct. What [tex]\lim_{n \to \infty} a_n = 0[/tex]means is that given [itex]\varepsilon_1 > 0[/itex], there exists [itex]N \in \mathbb{N}[/itex] such that n>N implies [itex]|a_n|\lt \varepsilon_1[/itex].

    Think about how you might relate [itex]\epsilon[/itex] for the anbn sequence to [itex]\varepsilon_1[/itex] to get what you need for the proof.
     
  6. Sep 2, 2011 #5
    You must show that, given any [itex]\epsilon > 0[/itex] that there exists an N > 0 such that if n> N, then [itex]|a_nb_n| < \epsilon[/itex]. Now you know that [itex](a_n)[/itex] converges to 0. That implies that, given any [itex]\epsilon > 0[/itex], then there exists an N > 0 such that [itex]|a_n| < \epsilon[/itex] if n > N. Does that N work? No, because if n > N then [itex]|a_nb_n| = |a_n| |b_n| < \epsilon M[/itex], but we needed [itex]|a_nb_n| < \epsilon[/itex]. Can you find a way to modify this argument to get the desired result? Perhaps your text or lecture notes contain examples that might help.
     
  7. Sep 2, 2011 #6
    That implies that, given any [itex]\epsilon > 0[/itex], then there exists an N > 0 such that [itex]|a_n| < \epsilon[/itex] if n > N. Does that N work? No, because if n > N then [itex]|a_nb_n| = |a_n| |b_n| < \epsilon M[/itex], but we needed [itex]|a_nb_n| < \epsilon[/itex].

    Could you just define [itex]\epsilon[/itex]M to be the new [itex]\epsilon[/itex]?

    Or can you say let [itex]\epsilon[/itex]>0, Since M is a real number>0 we know [itex]\epsilon[/itex]/M>0 so [itex]|a_nb_n| = |a_n| |b_n| < (\epsilon/M) *M=\epsilon[/itex],
     
    Last edited: Sep 2, 2011
  8. Sep 3, 2011 #7
    The last sentence in your post is basically correct. You just have to phrase it differently. Since you have the right idea, here's the way I would phrase it:

    Let [itex]\epsilon[/itex] be any number > 0. We have to show that there exists an N > 0 such that if n > N, then [itex]|a_nb_n| < \epsilon[/itex]. Now, since [itex](a_n)[/itex] converges to 0, there exists an [itex]N_1 > 0[/itex] such that if [itex]n > N_1[/itex], then [itex]|a_n| < \epsilon/M[/itex]. Therefore, if [itex]n > N_1[/itex], then [itex]|a_nb_n| < (\epsilon/M)(M) = \epsilon[/itex], as required.

    The point is that we define a different "[itex]\epsilon[/itex]" for [itex]a_n[/itex] so that we get [itex]|a_nb_n| < \epsilon[/itex]. Hope that makes sense.
     
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