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Convergent Series Question

  1. Sep 8, 2007 #1
    Hello, I have a question i can figure out.

    THE QUESTION:

    Show that

    [tex]\sqrt{2}, \sqrt{2\sqrt{2}}, \sqrt{2\sqrt{2\sqrt{2}}}, ...[/tex]

    converges and find the limit.


    From what I see, the first term is root 2, the second term will be the root of 2 times the first term making it larger than the first term, the third term will be the root of 2 times the second term making it larger than the second term.

    How can this converge to a real number?
     
  2. jcsd
  3. Sep 8, 2007 #2

    Gib Z

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    Homework Helper

    A formal proof would involve showing that the sequence [tex] x_n= \sqrt{2x_{n-1}} , x_1= \sqrt{2}[/tex] is strictly increasing and is bounded by a certain value, but really thats boring isn't it?

    Heres a less rigorous but more fun way :)

    Lets take some terms in the sequence and solve for 2 (Not that hard, just square both sides and divide both sides by 2 until the RHS is prepared :) :

    [tex]x_1^2 = 2[/tex]
    [tex] \frac{ x_2^4}{4} = 2[/tex]
    [tex] \frac{ x_3^8}{64} = 2[/tex]

    Ooo i Think i see a pattern :) Write the Denominators on the LHS as a power of 2, and take them over to the RHS :)

    [tex]x_3^8 = 2^7[/tex]

    We can see in general :
    [tex] x_n^{2^n} = 2^{2^n -1} [/tex]

    So using this for the 5th term,
    [tex] x_5^{32} = 2^{31}[/tex], which i've checked is true.

    Now, Take the log base 2 of the general form and use the following log rule, where b can be any base: [itex] \log_b y^k = k \log_b y[/itex].

    So we have: [tex]2^n \log_2 x_n = 2^n -1[/tex], dividing both sides by [itex]2^n[/tex] and then making both sides the exponents of 2 finally gives us:

    [tex] x_n = 2^{ \frac{2^n -1}{2^n}}[/tex].

    So now, take the limit as n approaches infinity, the exponent on the RHS becomes 1. ie
    [tex] \lim_{n\to \infty} x_n = 2[/tex].

    Very nice?
     
  4. Sep 8, 2007 #3

    Sir, thank you.
     
  5. Sep 8, 2007 #4

    Gib Z

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    Homework Helper

    No, thank YOU, I think thats the first time i've ever been called sir :D
     
  6. Sep 8, 2007 #5

    Gib Z

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    The monotone convergence theorem is the formal proof I mentioned you could do in my original post, Wikipedia states: If a_k is a monotone sequence of real numbers (e.g., if a_k ≤ a_{k+1},) then this sequence has a limit (if we admit plus and minus infinity as possible limits.) The limit is finite if and only if the sequence is bounded. I don't know why or how they could implement summation into it. Please post the proofs if the book included them.

    Edit: It seems you deleted your last post, making me look like a crackpot talking to myself lol
     
    Last edited: Sep 8, 2007
  7. Sep 8, 2007 #6
    No, no proof. the previous ones i did all utilized summation and that was the only technique they showed, but I'll tinker with it until tomorrow if i can't get it I'll surely post.


    Just trying to keep you on your toes.

    ;)
     
    Last edited: Sep 8, 2007
  8. Sep 11, 2007 #7
    Another Question

    I have another question for you.

    I have a sequence defined by:

    [tex]x_1=2[/tex]

    [tex]x_{n+1}=\frac{1}{2}(x_n + \frac{2}{x_n})[/tex]

    and i have to show that

    [tex]x^2_n[/tex]

    is always greater than 2.


    The only thing i can come up is this; I end up using two cases (its convoluted, its confusing, i don't even know if it qualifies as a proof, i don't know if its the best proof if it is a proof, but anyway) :

    i have to show that:

    [tex]x^2_n > 2[/tex] which is equivalent to [tex]x_n > \sqrt{2}[/tex]


    we assume

    [tex]x_n > \sqrt{2} [/tex].

    we know

    2 > [tex]\sqrt{2}[/tex].


    the expression

    [tex]x_n + 2[/tex]

    has two numbers > [tex]\sqrt{2}[/tex].


    when we add 2 to [tex]x_n[/tex] the result will either be a number twice as great as [tex]\sqrt{2}[/tex] or not.


    case 1:
    suppose adding 2 to [tex]x_n[/tex] does produce a number twice as great as [tex]\sqrt{2}[/tex].

    we know

    [tex]x^2_n > x_n > \sqrt{2}[/tex]

    so

    [tex]\frac{x^2_n}{x_n} > \sqrt{2}[/tex].

    but if

    [tex]\frac{x^2_n}{2x_n} < \sqrt{2}[/tex]

    then

    [tex]\frac{x^2_n + 2}{2x_n} > \sqrt{2}[/tex]


    case 2:
    suppose adding 2 to [tex]x_n[/tex] does not produce a number twice as great as [tex]\sqrt{2}[/tex].

    then we know [tex]x^2_n[/tex] is a number at least 4 times greater than [tex]\sqrt{2}[/tex].

    so

    [tex]\frac{x^2_n}{2x_n}[/tex] is still greater than [tex]\sqrt{2}[/tex].


    does that work as a proof?

    (by the way, i've still only been given Monotone Convergence and Cauchy Condensation to use to prove.)
     
    Last edited: Sep 11, 2007
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