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Math Help - using induction to prove...

  1. #1
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    using induction to prove...

    Hey everyone i was doing some AP work on my own and couldnt figure out how to do this question with induction. i know the steps just cant figure out how to use it in this case. Any help would be appreciated

    <br />
a(n) = \sqrt(2 +\sqrt 2+...+ \sqrt2)<br />
    "n roots"

    not sure if i typed it properly, but all of the square roots are under the first one so its the square root of (2 + squareroot of 2 + squareroot of 2) "n" times...

    then i have to:

    use induction to prove the sequence is increasing
    and prove if its bounded above by 2

    thanks in advance
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  2. #2
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    Quote Originally Posted by IrrationalPI3 View Post
    Hey everyone i was doing some AP work on my own and couldnt figure out how to do this question with induction. i know the steps just cant figure out how to use it in this case. Any help would be appreciated

    <br />
a_n = \sqrt{2 +\sqrt {2+ \sqrt{2+\cdots + \sqrt2}}}<br />
    "n roots"

    not sure if i typed it properly, but all of the square roots are under the first one so its the square root of (2 + squareroot of 2 + squareroot of 2) "n" times...

    then i have to:

    use induction to prove the sequence is increasing
    and prove if its bounded above by 2

    thanks in advance
    your sequence can be written as a_1=\sqrt{2} and a_{n+1}=\sqrt{2+a_n}, \ n \geq 1. now an easy induction over n shows that 0<a_n<2. then it follows that the sequence \{a_n \} is increasing because

    a_{n+1}=\sqrt{2 + a_n} > a_n is equivalent to (a_n + 1)(a_n - 2) < 0, which is obviously true because 0 < a_n < 2.
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  3. #3
    MHF Contributor red_dog's Avatar
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    a_n=\sqrt{2+\sqrt{2+\ldots+\sqrt{2}}}

    We can observe that a_n=\sqrt{2+a_{n-1}}, \ \forall n\geq 2

    a_1=\sqrt{2}, \ a_2=\sqrt{2+\sqrt{2}}\Rightarrow a_1<a_2

    Suppose that a_{n-1}<a_n

    a_n-a_{n+1}=\sqrt{2+a_{n-1}}-\sqrt{2+a_n}=\frac{a_{n-1}-a_n}{\sqrt{2+a_{n-1}}+\sqrt{2+a_n}}<0

    Then a_n<a_{n+1}


    a_1=\sqrt{2}<2

    Suppose that a_n<2.

    Then a_{n+1}=\sqrt{2+a_n}<\sqrt{2+2}=\sqrt{4}=2
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