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Math Help - Limit of sqrt(2*sqrt(2)) sequence

  1. #1
    s3a
    s3a is offline
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    Limit of sqrt(2*sqrt(2)) sequence

    I'm sure you people know which sequence I'm talking about. My teacher said it was popular. I'm just wondering if someone knows how to find the answer for this without just plugging in values with the calculator which is all I find myself being able to do with this problem.

    Any input would be appreciated!
    Thanks in advance!

    P.S. The limit (=answer) is 2.
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  2. #2
    MHF Contributor chiph588@'s Avatar
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    Quote Originally Posted by s3a View Post
    I'm sure you people know which sequence I'm talking about. My teacher said it was popular. I'm just wondering if someone knows how to find the answer for this without just plugging in values with the calculator which is all I find myself being able to do with this problem.

    Any input would be appreciated!
    Thanks in advance!

    P.S. The limit (=answer) is 2.
    Let  a_1=\sqrt2,\;a_2=\sqrt{2\sqrt2},\;\cdots,\; a_n=\underbrace{\sqrt{2\sqrt{2...\sqrt2}}}_{n}

    We then get  a_{n+1}=\underbrace{\sqrt{2\sqrt{2...\sqrt2}}}_{n+  1} = \sqrt{2a_n} or in other words  a_{n+1}^2=2a_n

    So if  L is our limit then it satisfies  L^2=2L \implies L=0\text{ or }2 .

    Show  \{a_n\}_{n=1}^\infty is an increasing sequence and that'll tell us  L\neq 0\implies L=2 .
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  3. #3
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    Alternatively,

    \sqrt{2\sqrt{2\sqrt{2...}}}=2^{\frac{1}{2}}2^{\fra  c{1}{4}}2^{\frac{1}{8}}....

    =2^{(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}....)}

    In brackets is a geometric series, first term=0.5, ratio=0.5.

    Limit=2^{S_{\infty}}=2^{(\frac{0.5}{1-0.5})}=2^{\frac{0.5}{0.5}}=2^1=2
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  4. #4
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    chiph588@'s method assume that sequence has a limit. Of course that can be proven by showing that the sequence is increasing and has 2 as an upper bound.
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