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Math Help - [SOLVED] Series

  1. #1
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    [SOLVED] Series

    Let x_n be the sequence with x_1=2 and x_n=\sqrt{5(x_{n-1})+6} for every integer n\geq 2. How can this be written in summation notation?

    x_1=2
    x_2=\sqrt{16}=4
    x_3=\sqrt{26}
    x_4=\sqrt{5\big(\sqrt{26}\big)+6}
    x_5=\sqrt{5\bigg(\sqrt{5\big(\sqrt{26}\big)+6}\big  g)+6}
    ......
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  2. #2
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    Quote Originally Posted by dwsmith View Post
    Let x_n be the sequence with x_1=2 and x_n=\sqrt{5(x_{n-1})+6} for every integer n\geq 2. How can this be written in summation notation?

    x_1=2
    x_2=\sqrt{16}=4
    x_3=\sqrt{26}
    x_4=\sqrt{5\big(\sqrt{26}\big)+6}
    x_5=\sqrt{5\bigg(\sqrt{5\big(\sqrt{26}\big)+6}\big  g)+6}
    ......
    You shouldn't be writing it with summation notation unless you intend to sum the terms of the sequence.
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    Quote Originally Posted by Prove It View Post
    You shouldn't be writing it with summation notation unless you intend to sum the terms of the sequence.
    That is what I intend to do.
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  4. #4
    Super Member Failure's Avatar
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    Quote Originally Posted by dwsmith View Post
    That is what I intend to do.
    What sum are you talking about? The x_n terms aren't sums at all: they are, obviously, square roots. And if you are talking about \sum_{n=1}^\infty x_n it is obvious that, since \lim_{n\to \infty}x_n=6, you have that \sum_{n=1}^\infty x_n=+\infty.

    Maybe what you are really interested in is \lim_{n\to\infty}x_n?
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  5. #5
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    Quote Originally Posted by Failure View Post
    What sum are you talking about? The x_n terms aren't sums at all: they are, obviously, square roots. And if you are talking about \sum_{n=1}^\infty x_n it is obvious that, since \lim_{n\to \infty}x_n=6, you have that \sum_{n=1}^\infty x_n=+\infty.

    Maybe what you are really interested in is \lim_{n\to\infty}x_n?
    Why is the limit 6?
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  6. #6
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    Quote Originally Posted by dwsmith View Post
    Why is the limit 6?
    Well, if the sequence x_n := \sqrt{5x_{n-1}+6} does have a limit, x_\infty, say, then, by continuity we should have that

    x_\infty = \lim_{n\to \infty}x_n=\lim_{n\to\infty}\sqrt{5x_{n-1}+6}=\sqrt{5\lim_{n\to\infty}x_{n-1}+6}=\sqrt{5x_\infty+6}

    But the equation x_\infty=\sqrt{5x_\infty+6} has only one solution, namely x_\infty=6.

    Of course, this does not prove that the sequence, starting with x_1=2, actually does converge to 6, but it shows that if the sequence converges at all, it converges to 6.

    That the sequence really does converge follows, for example, from Banch's Fixed Point Theorem (aka. Contraction Mapping Theorem), since f:\; [4;15]\to [4;15], x\mapsto \sqrt{5x+6} is contracting.
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