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Math Help - Proving an iterative sequence converges to a limit

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    Proving an iterative sequence converges to a limit

    The sequence is: a(n+1)=4+2(a(n))^(1/3) with a(1)=1

    Computing this clearly shows the sequence converges to 8, but how do I prove this?

    In earlier parts I have already proved the sequence is bounded, and increasing, so that proves that the sequence does converge, but now I need to prove what the limit is.

    Thanks
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    Quote Originally Posted by MickHarford View Post
    The sequence is: a(n+1)=4+2(a(n))^(1/3) with a(1)=1
    Computing this clearly shows the sequence converges to 8, but how do I prove this? In earlier parts I have already proved the sequence is bounded, and increasing, so that proves that the sequence does converge, but now I need to prove what the limit is.
    Can you solve L = 4 + 2\sqrt[3]{L}~?

    You see a_n~\&~a_{n+1} have the same limit L.
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    Great. Thanks, brings me back to college stuff I already know now!
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    To prove that it converges to 8, you will first need to prove it converges. After that, you can use Plato's idea. But before you can call the limit "L", you must show that the limit exists.

    For this problem and "iterative" problems in general, it is typically easiest to use the "monotone convergence theorem": "If a sequence of real number is increasing and has an upper bound, it converges".

    So we need to prove two things: that this sequence is has an upper bound and that it is increasing. Since we want this (increasing) sequence to converge to 8, a good candidate for an upper bound is 8.

    First, it is clear that all [itex]a_n[/itex] are positive.

    Prove, for all n, a_n< 8.
    And, again because this is "iterative", it is easiest to use 'proof by induction'. When i= 1, ]a_1= 1< 8 so the base case is true. Assume a_k< 8 for some k and look at k+1}. a_{k+1}= 4+ 2a_k^{1/3}. Because 0< a_k< 8, a_{k}^{1/3}< 8^{1/3}= 2. I'll let you finish this.

    Prove, for all n, a_{n+1}> a_n.
    We have a_1= 1 and [tex]a_2= 4+ 2(1^{1/2})= 4+ 2= 6> 1[/itex] so the base case is true.

    Assume that a_{k+1}> a_k for some k. Then a_{k+2}= 4+ 2(a_{k+1})^{1/3}. Because [itex]0< a_k< a_{k+1}[/itex], [itex]0< a_k^{1/3}< a_{k+1}^{1/3}[/itex]. Again, it should be easy to finish this.
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    I think it is worth noting that in the OP we have:
    Quote Originally Posted by MickHarford View Post
    The sequence is: a(n+1)=4+2(a(n))^(1/3) with a(1)=1
    Computing this clearly shows the sequence converges to 8, but how do I prove this?
    In earlier parts I have already proved the sequence is bounded, and increasing, so that proves that the sequence does converge,
    but now I need to prove what the limit is.
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    MHF Contributor chisigma's Avatar
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    The recursive relation can be written as...

    \displaystyle \Delta_{n}= a_{n+1}-a_{n} = 4 + 2\ a_{n}^{\frac{1}{3}} - a_{n}= f(a_{n}) (1)

    The function f(*) is illustrated here...



    There is only one 'attractive fixed point' in x_{0}=8 and, because \forall x is...

    \displaystyle |f(x)|<|x_{0}-x| (2)

    ... any a_{1} will produce a sequence that monotonically converges at x_{0}. The 'rigorous' demonstration of that is obtained considering that the sequence convergences at x_{0} if and only if converges the series...

    \displaystyle a_{1}+ \sum_{n=1}^{\infty} \Delta_{n} (3)

    Now it is easy to see that if (2) is satisfied, then the ratio test extablishes that (3) converges...

    Kind regards

    \chi \sigma
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