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Math Help - Limit of a recursively defined sequence

  1. #1
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    Limit of a recursively defined sequence

    I'm having some trouble with this question.

    Suppose that a_1=1 and a_{n+1}=\sqrt{1+a_n} whenever n\geq 1

    1) Show that \sqrt{1+x}\in [1,2] whenever x\in [1,2].


    2) Use induction to show that the sequence \{a_n\} is bounded.

    Would I do the question like this:

    Prove that a_n<2 \ \ \ \forall n\geq 1

    a_1=1<2

    Assume a_k<2, then a_{k+1}=\sqrt{1+a_k}<\sqrt{3}<2

    By induction it follows that a_n<2 \ \ \ \forall n\geq 1 \Rightarrow a_n is bounded above by 2.



    3) Use induction to show that {a_n} is an increasing sequence

    Prove a_{n+1}>a_n \ \ \ \forall n\geq 1

    a_2=\sqrt{1+1}=\sqrt{2}>a_1=1

    Assume that a_{k+1}>a_k

    a_{k+2}=\sqrt{1+a_{k+1}}>\sqrt{1+a_k}=a_{k+1}

    By induction it follows that, a_{n+1}>a_n \ \ \ \forall n\geq 1 \Rightarrow a_n is an increasing sequence



    4) Explain why \lim_{x \to \infty}a_n exists

    Since a_n is a monotonically increasing sequence, bounded above, it must converge to a limit as n\rightarrow\infty



    5) Find \lim_{x \to \infty}a_n. (That is, find \sqrt{1+\sqrt{1+\sqrt{1+...}}}

    Can someone check whether what I did is correct and help me with (1) and (5)
    Last edited by acevipa; September 13th 2010 at 04:03 AM.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by acevipa View Post
    I'm having some trouble with this question.

    Suppose that a_1=1 and a_{n+1}=\sqrt{1+a_n} whenever n\geq 1

    1) Show that \sqrt{1+x}\in [1,2] whenever x\in [1,2].

    Can someone check whether what I did is correct and help me with (1) and (5)
    f(x)=\sqrt{1+x} is increasing for x \in [0,\infty) , so we hace:

    1\le f(x)\le \sqrt{3}<2 for x\in [1,2]

    CB
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  3. #3
    Grand Panjandrum
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    Quote Originally Posted by acevipa View Post
    5) Find \lim_{x \to \infty}a_n. (That is, find \sqrt{1+\sqrt{1+\sqrt{1+...}}}

    Can someone check whether what I did is correct and help me with (1) and (5)
    You have shown that:

    a=\lim_{x \to \infty}a_n

    exists, and so is a solution of a=\sqrt{1+a}

    CB
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  4. #4
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    Quote Originally Posted by CaptainBlack View Post
    You have shown that:

    a=\lim_{x \to \infty}a_n

    exists, and so is a solution of a=\sqrt{1+a}

    CB
    I managed to work out (5)

    \lim_{n \to \infty}a_{n+1}=\lim_{n \to \infty}\sqrt{1+a_n}\Rightarrow a=\sqrt{1+a}\Rightarrow a^2=1+a

    \Rightarrow a^2-a-1=0\Rightarrow a=\frac{1\pm\sqrt{1+4}}{2}

    But a_n>0, so \lim_{n\to\infty}a_n=\frac{1+\sqrt{5}}{2}

    By the way, is my question 2, 3, 4 correct?
    Last edited by acevipa; September 13th 2010 at 04:38 AM.
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  5. #5
    Grand Panjandrum
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    Quote Originally Posted by acevipa View Post
    I don't quite understand what you did.

    The answer in the back of the book is \frac{1+\sqrt{5}}{2}

    By the way, is my question 2, 3, 4 correct?
    2, 3 and 4 look OK, but could use some tidying up.

    what I did was to observe that:

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

    implies that (since the limit exists):

    \lim a_{n+1}=\sqrt{1+\lim a_n}

    or:

    a=\sqrt{1+a}

    CB
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