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Math Help - Finding the Limit of a Sequence

  1. #1
    Junior Member Coop's Avatar
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    Question Finding the Limit of a Sequence

    Hi, I know the answer to the limit of this sequence but I am still unsure about the logic behind it.

    So I am given a sequence sqrt(6), sqrt(6+sqrt(6)), sqrt(6+sqrt(6+sqrt6)))...

    I know it is monotonic sequence, because it only increases.

    So a(n+1)=sqrt(6+a(n))

    And I can say that the lim(a(n+1))=lim(a(n)), right? Does that have to do it with it being monotonic?

    Here is where I am confused, an answer key says to denote the limit as something, say L and put it into the equation

    L = sqrt(6+L) (since the limits are equal)

    Then you get L^2 - L - 6 = 0

    And you get 3 as your answer, but why can you just put the limit (L) into the equation like that?

    Thanks

    P.S. If you don't get what my question is, I'd still appreciate your explanation on how to solve it.
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  2. #2
    MHF Contributor

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    Re: Finding the Limit of a Sequence

    " a_n" and " a_{n+1}" refer to exactly the same sequence, just renumbered. For example, a_n= 1/n, gives the sequence 1, 1/2, 1/3, 1/4, ... The sequence a_n= 1/(n+1) gives 1/2, 1/3, 1/4, ..., exactly the same except for the missing "1". And that doesn't change the limit, 0.

    So you have a_{n+1}= \sqrt{6+ a_n}. Since square root is continuous, taking the limit on both sides, lim_{n\to\infty} a_{n+1}= \sqrt{6+ lim_{n\to\infty} a_n. If we call the limit A, A must satisfy A= \sqrt{6+ A}. That's the equation you have.

    Now, that is assuming the sequence has limit. You say you know that the sequence is increasing. If you can show it has an upper bound, then you know it converges.
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  3. #3
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    Re: Finding the Limit of a Sequence

    Quote Originally Posted by Coop View Post
    Hi, I know the answer to the limit of this sequence but I am still unsure about the logic behind it.
    So I am given a sequence sqrt(6), sqrt(6+sqrt(6)), sqrt(6+sqrt(6+sqrt6)))...
    I know it is monotonic sequence, because it only increases.
    So a(n+1)=sqrt(6+a(n))
    And I can say that the lim(a(n+1))=lim(a(n)), right? Does that have to do it with it being monotonic?
    Here is where I am confused, an answer key says to denote the limit as something, say L and put it into the equation
    L = sqrt(6+L) (since the limits are equal)
    Then you get L^2 - L - 6 = 0
    And you get 3 as your answer, but why can you just put the limit (L) into the equation like that?
    Just saying it is increasing is not enough. It must also be bounded.
    But that is easy. You can show that \sqrt{6+a_n}\le 3

    If (a_n)\to L then also (a_{n+1})\to L.
    So if a_{n+1}=\sqrt{6+a_n} we have L=\sqrt{6+L}
    Thanks from Coop
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  4. #4
    Junior Member Coop's Avatar
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    Re: Finding the Limit of a Sequence

    Thanks Halls and Plato, this may be a trivial question but I am blanking on it; how would you go about proving that the sequence has a bound before you calculate the limit? Is the bound just [0,infinity), because square roots can't be negative? Or do you prove that the sequence has a bound through finding the limit?
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  5. #5
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    Re: Finding the Limit of a Sequence

    Quote Originally Posted by Coop View Post
    Thanks Halls and Plato, this may be a trivial question but I am blanking on it; how would you go about proving that the sequence has a bound before you calculate the limit?
    You know that a_1=\sqrt6<3. Now a_2=\sqrt{6+a_1}<\sqrt{6+3}=3

    By induction show a_n<3.
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  6. #6
    Junior Member Coop's Avatar
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    Re: Finding the Limit of a Sequence

    Oh okay, thank you
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