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Math Help - Proof of Convergence for Recursive Sequence

  1. #1
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    Proof of Convergence for Recursive Sequence

    I am trying to figure out how to prove that the following sequence converges:

    a_{n+1}=\sqrt{2+\sqrt{a_{n}}}

    with a_{1}=\sqrt{2}

    Thoughts anyone?
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  2. #2
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    Quote Originally Posted by General View Post
    Since the sequence converges, then \lim_{n\to\infty} a_n = L , where L is a finite number ..

    Recall that, for the convergent sequence a_n:
    \lim_{n\to\infty} a_n=\lim_{n\to\infty} a_{n+1} ..

    Since : a_{n+1}=\sqrt{2+\sqrt{a_{n}}}

    By taking the limit of both sides, we will conclude the following equation :

    L=\sqrt{2+\sqrt{L}} , solve it for L ..

    Thanks for the quick replies. I believe that in a formal proof

    L=\sqrt{2+\sqrt{L}}

    can be used to solve for the limit L if we know that the series converges. However, I need to first prove that the series converges given

    a_{1}=\sqrt{2}



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  3. #3
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    I think we should say this :

    |a_{n+1} - L|<\epsilon

    and  -\epsilon < a_{n+1} - L < \epsilon

    it means :

     0 < a_{n+1} - L < 2\epsilon

    and a_{n+1} < 2\epsilon+L

    also if 2\epsilon<<L we can unspot 2\epsilon .
    and a_{n+1} < L

    this is the proof. but to complete it




    we should proof that 2\epsilon<<L . or

    \epsilon<<\frac{L}{2}

    We Know L is 1 as here :

     \lim_{n\rightarrow\infty}\sqrt{\frac{2}{a_{n}} +1} = 1

    so the \epsilon<<0.5
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  4. #4
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    Quote Originally Posted by moemoe View Post
    However, I need to first prove that the series converges given
    Show that the sequence is increasing and bounded.
    Use induction to do that.
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  5. #5
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    Quote Originally Posted by parkhid View Post
    I think we should say this :

    |a_{n+1} - L|<\epsilon
    ...
    You can't assume convergence if you're trying to prove it...

    @moemoe: The usual way to prove that this type of recursive sequences converge is to prove by induction that the sequence is monotone and bounded.

    In your case, plug in the values for the first few terms to see whether the sequence is increasing or decreasing, and then prove it by induction. After you do that, prove that it is bounded (this can be done by induction as well).

    EDIT: Plato ahead of me :<
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  6. #6
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    So assuming

    <br />
a_{n} \geq a_{n-1} \geq 0<br />

    we have

    <br />
a_{n+1} = \sqrt{2+\sqrt{a_{n}}}  \geq \sqrt{2+\sqrt{a_{n-1}}}= a_{n}<br />

    Does anyone have a slick way of showing the sequence is bounded?
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  7. #7
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    Maybe define a new sequence

    <br />
b_{n+1}=\sqrt{2+\sqrt{b_{n}}}<br />
with <br />
b_{1}=2<br />

    where <br />
a_{n} \leq b_{n}<br />
for each n and <br />
b_{n}<br />
is decreasing by the same induction argument above?

    Is there another simpler way?
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  8. #8
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    There is a simpler way.

    First, validate that a_1, a_2 < 2.

    Now, assume a_n < 2 and use the induction hypothesis to prove that a_{n+1} < 2.
    This will prove that the sequence is bounded by 2 (it doesn't mean that 2 is the limit, though..)
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  9. #9
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    Perhaps we can try this.

    By the induction step it can be shown that a_{n}<2

    a_{n+1}=\sqrt{2+\sqrt{a_{n}}}

    (a_{n+1})^{2}=2+\sqrt{a_{n}}

    (a_{n+1})^{2}-1=1+\sqrt{a_{n}}

    (a_{n+1}+1)(a_{n+1}-1)=1+\sqrt{a_{n}}

    Since for all n, a_{n}<2

    a_{n+1}-1<1

    a_{n+1}+1>a_{n}+1

    a_{n+1}>a_{n}

    and it is monotone increasing.
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  10. #10
    MHF Contributor chisigma's Avatar
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    The sequence is defined recursively as...

    \Delta_{n} = a_{n+1} - a_{n} = \sqrt{2 + \sqrt{a_{n}}} - a_{n} = f(a_{n}) (1)

    The function f(x)= \sqrt{2 + \sqrt{x}} - x is represented here...



    It has a single zero at x_{0} = 1.83117720721... and because |f(x)| is less that the line crossing the x axis in x=x_{0} with unity negative slope, any a_{0} \ge 0 will produce a sequence converging at x_{0} without oscillations...

    Kind regards

    \chi \sigma
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