Results 1 to 8 of 8

Thread: sqrt(2)

  1. #1
    Junior Member
    Joined
    Sep 2007
    Posts
    46

    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.
    Follow Math Help Forum on Facebook and Google+

  2. #2
    MHF Contributor red_dog's Avatar
    Joined
    Jun 2007
    From
    Medgidia, Romania
    Posts
    1,252
    Thanks
    5
    $\displaystyle a_1=2^{\frac{1}{2}}$
    $\displaystyle a_2=2^{\frac{3}{4}}$
    $\displaystyle a_3=2^{\frac{7}{8}}$.
    By induction, $\displaystyle a_n=2^{\frac{2^n-1}{2^n}}$.
    Then $\displaystyle \lim_{n\to\infty}a_n=2$, so $\displaystyle a_n$ is convergent.
    Follow Math Help Forum on Facebook and Google+

  3. #3
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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: $\displaystyle a_1 = \sqrt{2}$ and $\displaystyle a_{n+1} = \sqrt{2a_n}$.
    Follow Math Help Forum on Facebook and Google+

  4. #4
    Senior Member
    Joined
    Apr 2006
    Posts
    401
    Quote Originally Posted by red_dog View Post
    $\displaystyle a_1=2^{\frac{1}{2}}$
    $\displaystyle a_2=2^{\frac{3}{4}}$
    $\displaystyle a_3=2^{\frac{7}{8}}$.
    By induction, $\displaystyle a_n=2^{\frac{2^n-1}{2^n}}$.
    Then $\displaystyle \lim_{n\to\infty}a_n=2$, so $\displaystyle a_n$ is convergent.
    So red_dog pretty much nailed it. I'll break it down for you.

    We have the sequence $\displaystyle x_n= \sqrt{2x_{n-1}} , x_1= \sqrt{2}$ 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:

    $\displaystyle
    x_1^2 = 2
    \frac{ x_2^4}{4} = 2
    \frac{ x_3^8}{64} = 2
    $

    See the pattern?

    $\displaystyle
    x_3^8 = 2^7
    $

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

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

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

    Take the limit as n goes to infinity:

    $\displaystyle \lim_{n\to \infty} x_n = 2 $
    Follow Math Help Forum on Facebook and Google+

  5. #5
    Junior Member
    Joined
    Sep 2007
    Posts
    46
    Quote Originally Posted by AfterShock View Post

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

    Take the limit as n goes to infinity:

    $\displaystyle \lim_{n\to \infty} x_n = 2 $
    How do you get that the lim is equal to 2 from $\displaystyle x_n = 2^{ \frac{2^n -1}{2^n}}$...
    Follow Math Help Forum on Facebook and Google+

  6. #6
    MHF Contributor

    Joined
    Aug 2006
    Posts
    21,739
    Thanks
    2812
    Awards
    1
    $\displaystyle \lim _{n \to \infty } \left( {\frac{{2^n - 1}}{{2^n }}} \right) = \lim _{n \to \infty } \left( {1 - \frac{1}{{2^n }}} \right) = 1\;\& \;2^1 = 2$
    Follow Math Help Forum on Facebook and Google+

  7. #7
    Junior Member
    Joined
    Sep 2007
    Posts
    46
    Quote Originally Posted by Plato View Post
    $\displaystyle \lim _{n \to \infty } \left( {\frac{{2^n - 1}}{{2^n }}} \right) = \lim _{n \to \infty } \left( {1 - \frac{1}{{2^n }}} \right) = 1\;\& \;2^1 = 2$
    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) ...
    Follow Math Help Forum on Facebook and Google+

  8. #8
    Global Moderator

    Joined
    Nov 2005
    From
    New York City
    Posts
    10,616
    Thanks
    10
    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 $\displaystyle f(x) = a^x$ for $\displaystyle a>0$ is a continous function. So if $\displaystyle c_n$ is a convergent sequence with limit $\displaystyle c$ then $\displaystyle a^{c_n}$ is a convergent sequence with limit $\displaystyle a^c$ by definition.
    Follow Math Help Forum on Facebook and Google+

Similar Math Help Forum Discussions

  1. Replies: 1
    Last Post: Nov 16th 2010, 04:54 AM
  2. Replies: 2
    Last Post: Sep 22nd 2010, 02:48 PM
  3. Prove Q(\sqrt{2}+\sqrt{3})=Q(\sqrt{2},\sqrt{3})
    Posted in the Advanced Algebra Forum
    Replies: 1
    Last Post: Apr 11th 2010, 04:52 PM
  4. Replies: 11
    Last Post: Jan 6th 2008, 09:33 AM
  5. prove sqrt(3) + sqrt (5) is an irrational number
    Posted in the Number Theory Forum
    Replies: 2
    Last Post: Oct 6th 2006, 06:48 PM

Search Tags


/mathhelpforum @mathhelpforum