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Math Help - Limit of infinite sequence.

  1. #1
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    Limit of infinite sequence.

    The sequence is for x>y>0
    <br />
a_1 = x +y,<br />
    <br />
a_n = x + y - \frac{xy}{a_{n-1}}<br />

    Now what i have got is
    <br />
a_n = \frac{x^{n+1}-y^{n+1}}{x^n - y^n}<br />

    How do i find the limit when n goes to infinity ?


    Edit: I suspect it has somthing to do with (I dont know the correct english word) the rule that when:
    <br />
(a_n) \leq (c_n) \leq (b_n)<br />

    <br />
 \lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} b_n<br />
    Then:
    <br />
\lim\limits_{n \to \infty} c_n = \lim\limits_{n \to \infty} a_n<br />
    But I am unable to construct two such limits.
    Last edited by hjortur; September 14th 2009 at 12:05 PM.
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  2. #2
    Super Member redsoxfan325's Avatar
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    Quote Originally Posted by hjortur View Post
    The sequence is for x>y>0
    <br />
a_1 = x +y,<br />
    <br />
a_n = x + y - \frac{xy}{a_{n-1}}<br />

    Now what i have got is
    <br />
a_n = \frac{x^{n+1}-y^{n+1}}{x^n - y^n}<br />

    How do i find the limit when n goes to infinity ?


    Edit: I suspect it has somthing to do with (I dont know the correct english word) the rule that when:
    <br />
(a_n) \leq (c_n) \leq (b_n)<br />

    <br />
 \lim\limits_{n \to \infty} a_n = \lim\limits_{n \to \infty} b_n<br />
    Then:
    <br />
\lim\limits_{n \to \infty} c_n = \lim\limits_{n \to \infty} a_n<br />
    But I am unable to construct two such limits.
    Do you know that it converges? If you do, then you know \{a_n\}\to a, so you can just take the limit of both sides now.

    \lim_{n\to\infty}a_n = \lim_{n\to\infty}\left(x+y-\frac{xy}{a_{n-1}}\right) \implies a = x+y-\frac{xy}{a}

    Just solve for a and you're done. If you don't know a_n converges, I would first try to prove that it does and then use the above method.
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  3. #3
    Super Member redsoxfan325's Avatar
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    Having tried out the sequence a few times with various x and y, it seems like the sequence converges to x, which is a solution of the above equation.
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  4. #4
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    I was able to find a:
    <br />
\lim_{n\to\infty}a_n = \lim_{n\to\infty}\left(x+y-\frac{xy}{a_{n-1}}\right) \implies a = x+y-\frac{xy}{a}<br />

    And the solution is a=x or a=y.
    The thing is I need to be able to show that it converges to x, and not y. How would I go about doing that?
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  5. #5
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    The limit is x, not y, because x> y. The initial value is x+y> x (because both x and y are positive) and the sequence decreases. The limit is x basically because it gets to x first!
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  6. #6
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    Jus shortly after writing this message I was able to show that the limit is x. Like i said, I knew the limit is x. But I needed to show that analyticaly.
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