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Math Help - Montone Recursive Sequence

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    Montone Recursive Sequence

    A sequence (x_n) is defined recursively as (x_n+1)=(6(x_n)+1)/(2(x_n)+5) with x1=3 for n є N
    Prove (x_n) is monotone

    By looking at the sequence it is decreasing in a monotone fashion but how do i construct a proof to show this?
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  2. #2
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    Quote Originally Posted by i_never_noticed View Post
    A sequence (x_n) is defined recursively as (x_n+1)=(6(x_n)+1)/(2(x_n)+5) with x1=3 for n є N
    Prove (x_n) is monotone

    By looking at the sequence it is decreasing in a monotone fashion but how do i construct a proof to show this?
    Perhaps you could try proof by induction.
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    i start that as (x_n+1)<(x_n)<3
    6(x_n+1)+1<6(x_n)+1<19
    but how do i then divided through is it correct in saying
    (6(x_n+1)+1)/(2(x_n+1)+5)<6(x_n)+1/(2(x_n)+5)<19/11??
    which would then be X4<X3<X2
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  4. #4
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    as 19/11 is X2
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    Quote Originally Posted by i_never_noticed View Post
    A sequence (x_n) is defined recursively as (x_n+1)=(6(x_n)+1)/(2(x_n)+5) with x1=3 for n є N
    Prove (x_n) is monotone

    By looking at the sequence it is decreasing in a monotone fashion but how do i construct a proof to show this?

    We form the difference::

    x_{n+1}-x_{n} = \frac{6x_{n} + 1}{2x_{n} +5}- x_{n} which after doing some calculations it is equal to:

    -\frac{(2x_{n} +1)(x_{n}-1)}{2x_{n} +5} and so we have:

    s_{n+1}-s_{n} =-\frac{(2x_{n} +1)(x_{n}-1)}{2x_{n} +5}.................................................. .........................................1

    Now we see that every term in (1),except x_{n}-1 ,is +ve because x_{n}>0 for all, n ,which we can prove easily by induction.

    So if we can prove x_{n}>1 for all n ,then (1) will be -ve and hence the sequence decreasing.

    And : x_{1}=3>1
    next suppose x_{n}>1\Longrightarrow [(6x_{n}+1>7)\wedge(2x_{n}+5>7)]\Longrightarrow\frac{6x_{n}+1}{2x_{n}+5}=x_{n+1}>1
    SO THE sequence is decreasing
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