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Math Help - Help

  1. #1
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    Question Help

    A sequence is defined by a1 = 1 and an = (2 + an-1) ^ (1/2) for n ≥ 2 Show that the sequence is increasing. Find the limit?
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  2. #2
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    Re: Help

    If you are not going to bother learning LaTeX, at least try to use the _ symbol to denote subscripts:

    If you wanted to type it into LaTeX, here is the code: [tex]a_1=1[/tex] and [tex]a_n = (2+a_{n-1})^{1/2}[/tex]

    Anyway, let f(x) = (2+x)^{1/2}. Then f'(x) = \dfrac{1}{2(2+x)^{1/2}}>0 for all -2 < x. So, f(x) is increasing showing that a_n>a_{n-1}. If the limit exists, then it must satisfy x = f(x). So, x = (2+x)^{1/2} implies x^2-x-2 = (x-2)(x+1) = 0, so x = 2 or x = -1. Since f(x) is increasing and f(1) = \sqrt{3}>1, so you know \lim_{n \to \infty} a_n \ge 1, so it must be \lim_{n \to \infty} a_n = 2.
    Last edited by SlipEternal; June 27th 2014 at 07:58 PM.
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  3. #3
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    Re: Help

    Tanks alot for the solution and fot the tip how to write the indeks, but you did mean f(2) in the solution not f(1)...


    Tanks again

    Best regards
    Haytham
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  4. #4
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    Re: Help

    Personally, I wouldn't use Calculus for a problem on sequences (although that is valid).

    Instead, use induction to prove " a_{n+1}> a_n for any integer n".

    We are given that a_1= 1 and a_2= \sqrt{2+ a_1}= \sqrt{2+ 1}= \sqrt{3}> 1 so the statement is true for n= 1.

    Assume that, for some integer, k, the statement is true: a_{k+1}> a_k. Then 2+ a_{k+1}> 2+ a_k and \sqrt{2+ a_{k+1}}> \sqrt{2+ a_k}. That is, a_{k+2}= \sqrt{2+ a_{k+1}}> \sqrt{2+ a_k}= a_{k+1} so the statement is true for k+1 as well. By induction, then, the statement is true for all n.

    To find the limit as n goes to infinity, first assume the limit exist- call it "A". Then applying the limit, as n goes to infinity to both sides of a_{n+1}= \sqrt{2+ a_n} we have A= \sqrt{2+ A} or A^2= 2+ A. A^2- A= 2. A^2- A+ \frac{1}{4}= (A- \frac{1}{2})^2= 2+ \frac{1}{4}= \frac{9}{4}.

    So A- \frac{1}{2}= \pm\frac{3}{2} and A= \frac{4}{2}= 2. I have taken the "+" because a_1= 1 and the sequence is increasing- the limit cannot be 1.

    That was, as I said, assuming the limit exists. We still need to prove that. Since the sequence is increasing we only need to prove that the sequence has an upper bound. Noticing that the sequence converged to 2, I think 2 would be a convenient upper bound. Prove by induction that " a_n< 2 for all n".

    a_1= 1< 2 so the statement is true for n= 1.

    Assume that a_k< 2 for some integer k. Then a_{k+1}= \sqrt{2+ a_k}< \sqrt{2+ 2}= \sqrt{4}= 2 and we are done. The sequence is increasing and has an upper bound, therefore it converges.
    Last edited by HallsofIvy; June 28th 2014 at 05:50 AM.
    Thanks from topsquark
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  5. #5
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    Re: Help

    Quote Originally Posted by Haytham1111 View Post
    Tanks alot for the solution and fot the tip how to write the indeks, but you did mean f(2) in the solution not f(1)...
    No, he didn't. Slip Eternal's f(x)= \sqrt{2+ x} is your a_{x+1}. a_2= f(1)

    Tanks again

    Best regards
    Haytham
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  6. #6
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    Re: Help

    Whats about A^2 - A + (1/4) = (A-(1/2))^2


    That is completely wrong
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  7. #7
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    Re: Help

    No, it's not. Multiply it yourself.
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  8. #8
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    Re: Help

    Quote Originally Posted by Haytham1111 View Post
    Whats about A^2 - A + (1/4) = (A-(1/2))^2


    That is completely wrong
    HallsofIvy completed the square rather than factoring or using the Quadratic Formula. Completing the square is actually the method used to prove the Quadratic Formula, so it is certainly a valid approach to solving any quadratic.
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  9. #9
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    Re: Help

    Ja, its right I understood it now. Thanks to you two
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