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Thread: Monotone and Cauchy Sequences

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    Monotone and Cauchy Sequences

    Let s1 = (6), s2 = (6+(6)), s3 = (6+(6+(6))), and in general define sn+1 = (6+sn). Prove that (sn) converges, and find its limit.

    *note: "√" means "the square root of"

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    Quote Originally Posted by bearej50 View Post
    Let s1 = (6), s2 = (6+(6)), s3 = (6+(6+(6))), and in general define sn+1 = (6+sn). Prove that (sn) converges, and find its limit.

    *note: "√" means "the square root of"

    show that the sequence is monotonic (increasing) and bounded. this will show that it has a limit (by what theorem?) you can prove both pretty easily by induction. do you see what to prove for each of these?

    to find the limit, note that $\displaystyle \lim s_{n + 1} = \lim s_n$, call this limit $\displaystyle L$.

    then, since $\displaystyle s_{n + 1} = \sqrt{6 + s_n}$ we have

    $\displaystyle \lim s_{n + 1} = \lim \sqrt{6 + s_n}$

    $\displaystyle \Rightarrow L = \sqrt{6 + L}$

    you want the positive solution (why?)
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    I am still confused...
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    is up to his old tricks again! Jhevon's Avatar
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    Quote Originally Posted by bearej50 View Post
    I am still confused...
    it would help if you say where specifically you are having trouble.

    take one part at a time. lets show that it is increasing. what have you done to this end?
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    I know that I have to prove that sn+1 > sn, which will prove that it is monotonically increasing. Then I would have to find an upper bound, U, for it and prove that sn < U and sn+1 < U. This would prove that it is bounded. Then, by the monotone convergence theorem, the sequence must be convergent and have a limit.
    I am having trouble putting this all together in a proper proof.
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    Here is the base case: $\displaystyle s_1 = \sqrt 6 < \sqrt {6 + \sqrt 6 } = s_2 < 3$.

    Say it is true for K:
    $\displaystyle \begin{gathered}
    s_K < s_{K + 1} < 3 \hfill \\
    6 + s_K < 6 + s_{K + 1} < 9 \hfill \\
    \underbrace {\sqrt {6 + s_K } }_{s_{K + 1} } < \underbrace {\sqrt {6 + s_{K + 1} } }_{s_{K + 2} } < 3 \hfill \\
    \end{gathered} $

    Done by induction: increasing and bounded.
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    Quote Originally Posted by bearej50 View Post
    I know that I have to prove that sn+1 > sn, which will prove that it is monotonically increasing. Then I would have to find an upper bound, U, for it and prove that sn < U and sn+1 < U. This would prove that it is bounded. Then, by the monotone convergence theorem, the sequence must be convergent and have a limit.
    I am having trouble putting this all together in a proper proof.
    O.K.


    Lets take the difference $\displaystyle S_{n+1}-S_{n} =\sqrt{6+s_{n}}-s_{n}$.................................................. ......................................1


    Now if you multiply (1) by $\displaystyle \frac{\sqrt{6+s_{n}}+s_{n}}{\sqrt{6+s_{n}}+s_{n}}$,and do couple of calculations you get the formula:

    $\displaystyle S_{n+1}-S_{n} =-\frac{(s_{n}+2)(s_{n}-3)}{\sqrt{6+s_{n}}+s_{n}}$.................................................. .....................................2

    In this formula all the terms are +ve except the factor $\displaystyle s_{n}-3$.

    But since we can prove that $\displaystyle s_{n}<3$ for all n, then by using (2) we get $\displaystyle S_{n+1}-S_{n}>0$.

    Hence the sequence is increasing and bounded from above by 3.

    So it is left to you to prove $\displaystyle s_{n}<3$ for all ,n and $\displaystyle s_{n}>0$ for all ,n.

    Up to now we have proved :


    a) IF the sequence converges ,then it must converge to 3 (because $\displaystyle s_{n}>0$ for all ,n)

    b) The sequence converges.

    BY (a) and (b) and using M.Ponens ,we have:

    The limit of the sequence is 3.

    Note usually the finding of a limit for this type of sequences consists of those two parts.

    In the first part WE Assume that the sequence converges,and find its limit .
    In the 2nd part we prove convergence
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