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Thread: Convergence Question

  1. #1
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    Convergence Question

    Consider the following sequence

    with a1 = and x>1.Show that it converges and that its limit is x.
    hint:Start by using induction to show that this sequence is non-decreasing and bounded above by x.
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  2. #2
    Grand Panjandrum
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    Quote Originally Posted by matty888 View Post
    Consider the following sequence

    with a1 = and x>1.Show that it converges and that its limit is x.
    hint:Start by using induction to show that this sequence is non-decreasing and bounded above by x.
    Take the hint, that shows the sequence converges. Then the limit satisfies the equation:

    $\displaystyle a=\sqrt{xa}$

    so

    $\displaystyle a^2-ax=0$

    hence $\displaystyle a=0$ or $\displaystyle a=x$. But the first of these is imposible, so $\displaystyle a=x$.

    RonL

    RonL
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  3. #3
    Super Member PaulRS's Avatar
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    Note that:
    • $\displaystyle a_{n}=x^{\frac{1}{2}}\cdot{a_{n-1}^{\frac{1}{2}}}$
    • $\displaystyle a_{n}=x^{\frac{1}{2}}\cdot{\left(x^{\frac{1}{2}}\c dot{a_{n-2}^{\frac{1}{2}}}\right)^{\frac{1}{2}}}$$\displaystyle =x^{\frac{1}{2}+\frac{1}{4}}\cdot{a_{n-2}^{\frac{1}{4}}}$
    • $\displaystyle a_n=x^{\frac{1}{2}+\frac{1}{4}+\frac{1}{8}}\cdot{a _{n-3}^{\frac{1}{8}}}$
    And so on.

    The general formula is: $\displaystyle
    a_n = x^{\sum\nolimits_{j = 1}^k {\left( {\tfrac{1}
    {2}} \right)^j } } \cdot a_{n - k} ^{{\raise0.5ex\hbox{$\scriptstyle 1$}
    \kern-0.1em/\kern-0.15em
    \lower0.25ex\hbox{$\scriptstyle {2^k }$}}}
    $ for $\displaystyle
    1 \leqslant k \leqslant n-1
    $

    We now let $\displaystyle k=n-1$ ==>$\displaystyle
    a_n = x^{\sum\nolimits_{j = 1}^{n - 1} {\left( {\tfrac{1}
    {2}} \right)^j } } \cdot a_1 ^{\tfrac{1}
    {{2^{n - 1} }}}


    $

    Thus: $\displaystyle
    \mathop {\lim }\limits_{n \to + \infty } a_n = \mathop {\lim }\limits_{n \to + \infty } x^{\sum\nolimits_{j = 1}^{n - 1} {\left( {\tfrac{1}
    {2}} \right)^j } } \cdot a_1 ^{\tfrac{1}
    {{2^{n - 1} }}}

    $

    Applying the Geometric Sum formula we get $\displaystyle
    a_n = x^{\sum\nolimits_{j = 1}^{n - 1} {\left( {\tfrac{1}
    {2}} \right)^j } } \cdot a_1 ^{\tfrac{1}
    {{2^{n - 1} }}} = x^{1 - \left( {\tfrac{1}
    {2}} \right)^n }

    $

    Therefore: $\displaystyle
    \mathop {\lim }\limits_{n \to + \infty } a_n = x
    $
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