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Thread: Investigating a Sequence

  1. #1
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    Investigating a Sequence

    The sequence is recursively defined by a1 = √(2), an+1 = √(2+an) for n ≥ 1.

    The sequence {an} is bounded above. Call this upper bound L.

    Show {an} is monotone increasing. Thus {an} has a limit. Why?

    Find this limit?

    Thank you sooo much for all the help guys!!!
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hi
    Quote Originally Posted by jules027 View Post
    The sequence is recursively defined by a1 = √(2), an+1 = √(2+an) for n ≥ 1.

    The sequence {an} is bounded above. Call this upper bound L.

    Show {an} is monotone increasing. Thus {an} has a limit. Why?

    Let $\displaystyle f:x\mapsto\sqrt{2+x}$.

    $\displaystyle a_{n+1}=\sqrt{2+a_n}=f(a_n)$ hence the sequence is monotone increasing iff $\displaystyle f$ is monotone increasing.
    Find this limit?
    The limit
    $\displaystyle l$ of the sequence is such that $\displaystyle f(l)=l$ because $\displaystyle f$ is continuous on $\displaystyle [\sqrt{2},\,L]$.
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  3. #3
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    Quote Originally Posted by jules027 View Post
    The sequence {an} is bounded above. Call this upper bound L.

    ...{an} is monotone increasing. Thus {an} has a limit. Why?


    You have by definition that a number L is the limit of a sequence $\displaystyle ({x_n:n\in\mathbb{N}})$ iff $\displaystyle \forall{\epsilon >0} \exists{N}\in\mathbb{N}:\forall{n>N}, {|{x_n}-L|}<\epsilon$

    A sequence converges iff it has a limit.

    You can see from the definition that every monotone bounded sequence has a limit and thus converges.

    So your sequence $\displaystyle (a_n)$ being both monotone and bounded, necessarily converges and thus has a limit.
    Last edited by TXGirl; May 17th 2008 at 09:43 AM.
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  4. #4
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    Quote Originally Posted by flyingsquirrel View Post
    Hi

    Let $\displaystyle f:x\mapsto\sqrt{2+x}$.

    $\displaystyle a_{n+1}=\sqrt{2+a_n}=f(a_n)$ hence the sequence is monotone increasing iff $\displaystyle f$ is monotone increasing.
    The limit $\displaystyle l$ of the sequence is such that $\displaystyle f(l)=l$ because $\displaystyle f$ is continuous on $\displaystyle [\sqrt{2},\,L]$.
    So there would be no computations to reach the limit? It is just the statement,
    "The limit $\displaystyle l$ of the sequence is such that $\displaystyle f(l)=l$ because $\displaystyle f$ is continuous on $\displaystyle [\sqrt{2},\,L]$."?
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  5. #5
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    Quote Originally Posted by jules027 View Post
    So there would be no computations to reach the limit? It is just the statement,
    "The limit $\displaystyle l$ of the sequence is such that $\displaystyle f(l)=l$ because $\displaystyle f$ is continuous on $\displaystyle [\sqrt{2},\,L]$."?
    Actually $\displaystyle f(a_n) = a_{n+1}$. If the limit of the sequence is L, then by letting n tend to infinity we get $\displaystyle f(L) = L$

    $\displaystyle \sqrt{2+L} = L \Rightarrow L^2 - L - 2 = 0$
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  6. #6
    Super Member flyingsquirrel's Avatar
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    Quote Originally Posted by jules027 View Post

    The sequence {an} is bounded above. Call this upper bound L.

    Quote Originally Posted by Isomorphism View Post
    If the limit of the sequence is L ...

    These two $\displaystyle L$ are not necessarily the same one.
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  7. #7
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    So how would I get at this limit exactly then? Thanks for the help!
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  8. #8
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    Quote Originally Posted by Isomorphism View Post
    Actually $\displaystyle f(a_n) = a_{n+1}$. If the limit of the sequence is L, then by letting n tend to infinity we get $\displaystyle f(L) = L$

    $\displaystyle \sqrt{2+L} = L \Rightarrow L^2 - L - 2 = 0$
    Quote Originally Posted by flyingsquirrel View Post
    These two $\displaystyle L$ are not necessarily the same one.
    flyingsquirrel means generally they need not be the same one. Since what I said could be misleading. You might think it works for any function.

    However for $\displaystyle a_{n+1} = \sqrt{2 + a_n}$, we can limit both sides to infinity to get $\displaystyle \lim_{n \to \infty} a_{n+1} =\lim_{n \to \infty} \sqrt{2 + a_n} = \sqrt{2 + \lim_{n \to \infty} a_n}$

    Since you have already proved that $\displaystyle (a_n)$ has a limit. Call it l. Its easy to see that $\displaystyle \lim_{n \to \infty} a_n = \lim_{n \to \infty} a_{n+1} = l$. So you can continue like what I did before.
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  9. #9
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    I am still confused about this one. Could somebody try and help? Maybe a step by step procedure of how to get at the answer? I almost have it but I am confused on the steps to get there. Thanks so much!!
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by Isomorphism View Post
    flyingsquirrel means generally they need not be the same one. Since what I said could be misleading. You might think it works for any function.

    However for $\displaystyle a_{n+1} = \sqrt{2 + a_n}$, we can limit both sides to infinity to get $\displaystyle \lim_{n \to \infty} a_{n+1} =\lim_{n \to \infty} \sqrt{2 + a_n} = \sqrt{2 + \lim_{n \to \infty} a_n}$

    Since you have already proved that $\displaystyle (a_n)$ has a limit. Call it l. Its easy to see that $\displaystyle \lim_{n \to \infty} a_n = \lim_{n \to \infty} a_{n+1} = l$. So you can continue like what I did before.
    As Isomorphism showed

    $\displaystyle \lim_{n\to\infty}a_{n+1}=\sqrt{2+\lim_{n\to\infty} }\Rightarrow{(\lim_{n\to\infty}a_{n+1})^2=2+\lim_{ n\to\infty}a_{n}}$

    and since as $\displaystyle n\to\infty$ $\displaystyle a_n=a_{n+1}$

    we say Let $\displaystyle a_n=\lambda$

    So we get

    $\displaystyle \lambda^2=2+\lambda\Rightarrow{\lambda^2-2-\lambda=0}$

    Now just solve for $\displaystyle \lambda$
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  11. #11
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    So the final limit would be from (-1,2)?
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  12. #12
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by jules027 View Post
    So the final limit would be from (-1,2)?
    No...Unless I am misreading your answer it should be just 2
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