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Math Help - proving the limit of a two sequences

  1. #1
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    proving the limit of a two sequences

    If a_n --> 0 and there is a sequence b_n that is bounded, then a_n * b_n --> 0

    I am lost on how to prove this.

    Since we only know that b_n is bounded, it doesn't necessarily converge or have a limit?
    So how do I go about proving that a_n * b_n has a limit?
    Thanks!
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  2. #2
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cubs3205 View Post
    If a_n --> 0 and there is a sequence b_n that is bounded, then a_n * b_n --> 0

    I am lost on how to prove this.

    Since we only know that b_n is bounded, it doesn't necessarily converge or have a limit?
    So how do I go about proving that a_n * b_n has a limit?
    Thanks!
    since b_n is bounded, there is some constant M > 0 so that |b_n| \le M for all n

    but that means |a_nb_n| = |a_n||b_n| \le M|a_n| \cdots
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  3. #3
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    Sorry, I am really trying, but I am not getting anywhere. What you provided me helped and made sense but I'm not sure how to continue.
    My actual proof experience (besides the stuff I tried on my own) only started three classes ago
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cubs3205 View Post
    Sorry, I am really trying, but I am not getting anywhere. What you provided me helped and made sense but I'm not sure how to continue.
    My actual proof experience (besides the stuff I tried on my own) only started three classes ago
    when proving something, if you have no idea where to start, just go by the definition. what does it mean for the limit to approach something? what is the definition of a limit? what is it that you have to show?

    you should know that we say the limit of a sequence \{ s_n \} exists and equals L iff

    For every \epsilon > 0, there is some N \in \mathbb{N} such that n > N implies |s_n - L| < \epsilon

    what can you do with that?


    when you get a bit more experienced with that approach, then you can use more sophisticated approaches, like the squeeze theorem, which is what i would use here. what can you say about \lim |a_nb_n| ?
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  5. #5
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    Ah with the squeeze theorem, then -M(a_n) < a_n * b_n < M(a_n)
    Since M(a_n) --> M * (limit of a_n) = 0
    Then a_n * b_n --> 0

    I think?
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  6. #6
    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by cubs3205 View Post
    Ah with the squeeze theorem, then -M(a_n) < a_n * b_n < M(a_n)
    Since M(a_n) --> M * (limit of a_n) = 0
    Then a_n * b_n --> 0

    I think?
    yeah, something like that. use \le signs though

    since \lim |a_nb_n| \le 0 (meaning 0 \le \lim a_nb_n \le 0), we have that \lim (a_nb_n) = 0 by the Squeeze Theorem
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