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Thread: convergent sequence

  1. #1
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    convergent sequence

    Suppose $\displaystyle x_1=\sqrt{2}$ and $\displaystyle x_{n+1}=\sqrt{2x_n}$ for$\displaystyle n\geq 1$. Show that the sequence converges.
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  2. #2
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    Quote Originally Posted by Kat-M View Post
    Suppose $\displaystyle x_1=\sqrt{2}$ and $\displaystyle x_{n+1}=\sqrt{2x_n}$ for$\displaystyle n\geq 1$. Show that the sequence converges.
    $\displaystyle x_1 < x_2 = \sqrt {2x_1 } = \sqrt {2\sqrt 2 } < 2$

    $\displaystyle \begin{gathered}
    x_{N - 1} < x_N < 2 \hfill \\
    2x_{N - 1} < 2x_N < 4 \hfill \\
    \sqrt {2x_{N - 1} } < \sqrt {2x_N } < \sqrt 4 \hfill \\
    x_N < x_{N + 1} < 2 \hfill \\
    \end{gathered} $

    So it increasing bounded above. Thus $\displaystyle \left( {x_N } \right) \to L\;\& \,L = \sqrt {2L} $
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