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Math Help - sqrt(2)

  1. #1
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    sqrt(2)

    Show that

    sqrt(2), sqrt(2*sqrt(2)), sqrt(2*sqrt(2*sqrt(2))), ...

    converges, and find the limit.

    --------------------------

    So, uh, to show that it converges, we can use the monotone convergence theorem, that says if its bounded and monotone, it will converge... maybe?

    EDIT:

    Oh, maybe use the theorem that says that subsequences of a convergent sequence converge to the same lim as the orig. sequence? I'm not too sure.
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  2. #2
    MHF Contributor red_dog's Avatar
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    a_1=2^{\frac{1}{2}}
    a_2=2^{\frac{3}{4}}
    a_3=2^{\frac{7}{8}}.
    By induction, a_n=2^{\frac{2^n-1}{2^n}}.
    Then \lim_{n\to\infty}a_n=2, so a_n is convergent.
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  3. #3
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    Quote Originally Posted by seerTneerGevoLI View Post
    Show that

    sqrt(2), sqrt(2*sqrt(2)), sqrt(2*sqrt(2*sqrt(2))), ...

    converges, and find the limit.
    .
    Hint: a_1 = \sqrt{2} and a_{n+1} = \sqrt{2a_n}.
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  4. #4
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    Quote Originally Posted by red_dog View Post
    a_1=2^{\frac{1}{2}}
    a_2=2^{\frac{3}{4}}
    a_3=2^{\frac{7}{8}}.
    By induction, a_n=2^{\frac{2^n-1}{2^n}}.
    Then \lim_{n\to\infty}a_n=2, so a_n is convergent.
    So red_dog pretty much nailed it. I'll break it down for you.

    We have the sequence x_n= \sqrt{2x_{n-1}} , x_1= \sqrt{2} which is strictly increasing and bounded (this would require a rigorous proof...but since you just said show, I'll show it in a simple way).

    We have:

    <br />
x_1^2 = 2<br />
\frac{ x_2^4}{4} = 2<br />
\frac{ x_3^8}{64} = 2<br />

    See the pattern?

    <br />
x_3^8 = 2^7<br />

    In general, x_n^{2^n} = 2^{2^n -1}

    Use the log rule:  \log_b y^k = k \log_b y

    We have:  2^n \log_2 x_n = 2^n -1. Divide both sides by  2^n and thus:  x_n = 2^{ \frac{2^n -1}{2^n}}.

    Take the limit as n goes to infinity:

    \lim_{n\to \infty} x_n = 2
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  5. #5
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    Quote Originally Posted by AfterShock View Post

    We have:  2^n \log_2 x_n = 2^n -1. Divide both sides by  2^n and thus:  x_n = 2^{ \frac{2^n -1}{2^n}}.

    Take the limit as n goes to infinity:

    \lim_{n\to \infty} x_n = 2
    How do you get that the lim is equal to 2 from  x_n = 2^{ \frac{2^n -1}{2^n}}...
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  6. #6
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    \lim _{n \to \infty } \left( {\frac{{2^n  - 1}}{{2^n }}} \right) = \lim _{n \to \infty } \left( {1 - \frac{1}{{2^n }}} \right) = 1\;\& \;2^1  = 2
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  7. #7
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    Quote Originally Posted by Plato View Post
    \lim _{n \to \infty } \left( {\frac{{2^n - 1}}{{2^n }}} \right) = \lim _{n \to \infty } \left( {1 - \frac{1}{{2^n }}} \right) = 1\;\& \;2^1 = 2
    Hmm, I discussed this with my professor, and he said something along the lines of there being no theorem stating something like:

    lim a^(c_n) = lim a^(lim c_n)

    EDIT:

    You're doing 2^(lim c_n) ...
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  8. #8
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    Quote Originally Posted by seerTneerGevoLI View Post
    Hmm, I discussed this with my professor, and he said something along the lines of there being no theorem stating something like:

    lim a^(c_n) = lim a^(lim c_n)

    EDIT:

    You're doing 2^(lim c_n) ...
    The function f(x) = a^x for a>0 is a continous function. So if c_n is a convergent sequence with limit c then a^{c_n} is a convergent sequence with limit a^c by definition.
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