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Thread: sequence questions

  1. #1
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    sequence questions

    (a) Prove that if 0<a<2, then a<2$\displaystyle \sqrt{a}$<2.

    (b) Prove that the sequence $\displaystyle \sqrt{2}$, $\displaystyle \sqrt{2\sqrt{2}}$, $\displaystyle \sqrt{2\sqrt{2\sqrt{2}}}$ converges.

    (c) Find the limit of the sequence.


    Thanks.
    Last edited by khatz; Jul 19th 2009 at 08:18 PM.
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  2. #2
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    Quote Originally Posted by khatz View Post
    (a) Prove that if 0<a<2, then a<2$\displaystyle \sqrt{2}$<2.
    It isn't true! Must be a typo somewhere.
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  3. #3
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    For (b) show that the sequence is increasing (I wont show that here) and bounded.

    Clearly $\displaystyle a_1=\sqrt{2}<2$ so assume that $\displaystyle a_n<2$ for some n. Then $\displaystyle a_{n+1}=\sqrt{2a_n}<\sqrt{2*2}=2$ so $\displaystyle a_n<2$ for all n and is thus bounded.
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  4. #4
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    Quote Originally Posted by AlephZero View Post
    It isn't true! Must be a typo somewhere.
    Yes, it should be 2sqrt(a).
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  5. #5
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    Quote Originally Posted by khatz View Post
    Yes, it should be 2sqrt(a).
    Still not true!

    Counterexample: $\displaystyle a=1.9 < 2 \implies \sqrt{a}\approx 1.38 \implies 2\sqrt{a} \approx 2.76 > 2.$
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  6. #6
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    Quote Originally Posted by khatz View Post
    (a) Prove that if 0<a<2, then a<2$\displaystyle \sqrt{a}$<2.

    (c) Find the limit of the sequence.


    Thanks.
    Considering what it leads onto, I think (a) should be

    (a) Prove that if $\displaystyle 0<a<2$, then $\displaystyle a< \sqrt{2a}<2$


    For (c), if the limit it exists we can say that $\displaystyle \sqrt{2 \sqrt{2\sqrt{2\sqrt{\ldots}}}} = x$

    So what can you say about $\displaystyle x^2$ ?
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