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Math Help - 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:

    a=\sqrt{xa}

    so

    a^2-ax=0

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

    RonL

    RonL
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  3. #3
    Super Member PaulRS's Avatar
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    Note that:
    • a_{n}=x^{\frac{1}{2}}\cdot{a_{n-1}^{\frac{1}{2}}}
    • a_{n}=x^{\frac{1}{2}}\cdot{\left(x^{\frac{1}{2}}\c  dot{a_{n-2}^{\frac{1}{2}}}\right)^{\frac{1}{2}}} =x^{\frac{1}{2}+\frac{1}{4}}\cdot{a_{n-2}^{\frac{1}{4}}}
    • 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: <br />
a_n  = x^{\sum\nolimits_{j = 1}^k {\left( {\tfrac{1}<br />
{2}} \right)^j } }  \cdot a_{n - k} ^{{\raise0.5ex\hbox{$\scriptstyle 1$}<br />
\kern-0.1em/\kern-0.15em<br />
\lower0.25ex\hbox{$\scriptstyle {2^k }$}}} <br />
for <br />
1 \leqslant k \leqslant n-1<br />

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

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

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

    Therefore: <br />
\mathop {\lim }\limits_{n \to  + \infty } a_n  = x<br />
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