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Math Help - Limits

  1. #1
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    Limits

    Hey, I'm stuck on this question and desperately need help.

    Question

    Suppose that the sequence (a_n) is convergent, and assume that the sequence (b_n) is bounded. Prove that the sequence (c_n) defined by

     c_n = \frac{a_nb_n - 5n}{a^2_n + 3n}

    is convergent and find its limit. (n belongs to the set of natural numbers)


    ================================================== =======

    Please help! I tried using the theorem that if a sequence is convergent, then it is bounded (for (a_n) ) but get anywhere. It's quite clear to me that the limit is \frac{-5}{3} but I don't know how to get there.
    Last edited by Joel24; January 24th 2008 at 01:30 PM.
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  2. #2
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    Quote Originally Posted by Joel24 View Post
    Hey, I'm stuck on this question and desperately need help.

    Question

    Suppose that the sequence (a_n) is convergent, and assume that the sequence (b_n) is bounded. Prove that the sequence (c_n) defined by

     c_n = \frac{a_nb_n - 5n}{a^2_n + 3n}
    Since c_n has a possibility is not being properly defined we should make a restriction like a_n,b_n\geq 0.


    Hint: Divide the numerator and denominator by n and take the limit.
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  3. #3
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    Quote Originally Posted by ThePerfectHacker View Post
    Since c_n has a possibility is not being properly defined we should make a restriction like a_n,b_n\geq 0.


    Hint: Divide the numerator and denominator by n and take the limit.
    Ok, thanks for the reply. Here's what I've done:
    __________________________________________________ _____________

     c_n = \frac{ \frac{a_n b_n}{n} - 5}{\frac{a^2_n}{n} + 3}

    Let  x \in \{ a_n : n is a natural number \}, and let  M_1 be an upper bound of the set.

    Let  y \in \{ b_n : n is a natural number \}, and let  M_2 be an upper bound of the set.

    Then, for all  a_n, b_n > 0

     \lim_{n \to \infty} \left( \frac{ \frac{xy}{n} - 5}{ \frac{x^2}{n} + 3} \right)

    But as  x \leq M_1 for all  x and y  y \leq M_2 for all  y , it follows that

     \lim_{n \to \infty} \left( \frac{xy \left( \frac{1}{n}\right) - 5}{x^2 \left( \frac{1}{n} \right) + 3} \right) = \frac{-5}{3}

    __________________________________________________ ______________

    But I don't know how to prove the sequence is convergent before finding the limit as the question asks.

    Could I perhaps show the sequence is monotone and bounded and therefore convergent? I've tried, but couldn't prove it.

    Also, what if  a_n < 0 and  b_n \geq 0 ? We made a restriction but surely this doesn't cover all possibilities.

    Sorry if I'm asking silly questions, but I really need help with this.
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  4. #4
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    Quote Originally Posted by Joel24 View Post
    Also, what if  a_n < 0 and  b_n \geq 0 ? We made a restriction but surely this doesn't cover all possibilities.

    Sorry if I'm asking silly questions, but I really need help with this.
    If we do not make this restrictions then the denominator can be zero and the sequences are not well-defined. It is absolutely necessary to make some sort of restriction.


    Just use the fact that is a_n is bounded then \frac{a_n}{n} \to 0. Thus, if a sequence a_n is convergent (therefore bounded) we have \frac{a_n}{n}\to 0.

    This means,
    \frac{\frac{a_nb_n}{n} - 5}{\frac{a_n^2}{n} + 3} \to -\frac{5}{3}

    Because a_nb_n is bounded and a_n^2 is bounded.
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  5. #5
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    Ok, thank you.

    In the seminars, we haven't been told that  lim_{n \to \infty} \left(\frac{a_n}{n} \right) = 0 if the sequence is bounded. As obvious as it is.

    So do you think it would be more 'rigorous' if I said that since  a_n is bounded, there exists an M such that  | a_n | \leq M

    Then I can show that  lim_{n \to \infty} \left(\frac{M}{n} \right) = 0 and  lim_{n \to \infty} \left(\frac{-M}{n} \right) = 0 . Since  -M \leq a_n \leq M,  \forall n then by the sandwich theorem,  lim_{n \to \infty} \left(\frac{a_n}{n} \right) = 0

    Is that fine, in your opinion, or unnecessary?

    Thanks.
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  6. #6
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    Quote Originally Posted by Joel24 View Post
    Ok, thank you.

    In the seminars, we haven't been told that  lim_{n \to \infty} \left(\frac{a_n}{n} \right) = 0 if the sequence is bounded. As obvious as it is.

    So do you think it would be more 'rigorous' if I said that since  a_n is bounded, there exists an M such that  | a_n | \leq M

    Then I can show that  lim_{n \to \infty} \left(\frac{M}{n} \right) = 0 and  lim_{n \to \infty} \left(\frac{-M}{n} \right) = 0 . Since  -M \leq a_n \leq M,  \forall n then by the sandwich theorem,  lim_{n \to \infty} \left(\frac{a_n}{n} \right) = 0
    That is excellent.
    Is that fine, in your opinion, or unnecessary?

    Thanks.
    That is excellent.
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