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Math Help - Unbounded sequence

  1. #1
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    Unbounded sequence

    Let x_{1} =1 , and let x_{n+1} =\frac {(x_1 +x_2 + . . . + x_{n} ) }{2} . Prove that x_n \rightarrow \infty

    So far:

    I have proved that  x_n is a strictly decreasing sequence by induction. Will this help a bit?
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  2. #2
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    Hello,
    Quote Originally Posted by tttcomrader View Post
    Let x_{1} =1 , and let x_{n+1} =\frac {(x_1 +x_2 + . . . + x_{n} ) }{2} . Prove that x_n \rightarrow \infty

    So far:

    I have proved that  x_n is a strictly decreasing sequence by induction. Will this help a bit?
    Uh... Don't you find it weird that you're asked to prove that x_n \to +\infty and though you find it to be decreasing ?

    Consider n \ge 2 and see if it's decreasing or increasing ^^
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  3. #3
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    I have x1=1, x2=1/2, x3=3/4, x4=9/8.

    so it is in fact increasing strictly...
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  4. #4
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    Can you show k \ge 2\quad  \Rightarrow \quad x_k  = \frac{{3^{k - 2} }}{{2^{k - 1} }}?
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  5. #5
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    It is true for k=2.

    If it is true for k=n, then we have x_k = \frac {x_1 + . . . + x_{k-1} }{2}= \frac {3^{k-2}}{2^{k-1}}

    Then  x_{k+1} = \frac {x_1 + . . . + x_{k-1} + x_{k} } {2} = \frac {3 ^ {k-2} }{2^{k-1}} + \frac {x_k}{2} =  \frac {3 ^ {k-2} }{2^{k-1}} + \frac {3^{k-2}}{2^{k}}

    = \frac { (2)3^{k-2} +3^{k-2} } {2^k} = \frac {3^ {k-1}}{2^k} = \frac {3^{(k+1)-2}} {2^{(k+1)-1}}

    So it is true for all  k \in \mathbb {N}

    Here is my proof to finish the problem:

    x_n = \frac {3^{k-2}}{2^{k-1}} = \frac {2}{3^{2}} ( \frac {3}{2})^k \geq \frac {2}{3^{2}} \ \ \ \ \ \ \forall k \in \mathbb {N}

    So this sequence is not bounded.

    Is this correct? Thank you.
    Last edited by tttcomrader; September 6th 2008 at 01:41 PM.
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