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Math Help - Induction on sequences

  1. #1
    Senior Member I-Think's Avatar
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    Induction on sequences

    The sequence of positive numbers u_1, u_2, u_3,... is such that u_1<4
    and
     u_{n+1}=\frac{5u_n+4}{u_n+2}

    By considering 4-u_{n+1}, prove by induction that u_n<4 for all n\geq{1}

    My attempt N.B To make this easier to read I'm skipping all the official statements of the induction process
    4-u_{n+1}=\frac{4-u_n}{2+u_n}

    Let n=1
    4-u_2=\frac{4-u_1}{2+u_1}
    It is given that u_1<4 and positive so \frac{4-u_1}{2+u_1}<4
    u_2=4-\frac{4-u_1}{2+u_1}
    Hence u_2 is less than 4

    Assume true for n=k
    4-u_{k+1}=\frac{4-u_k}{2+u_k}

    Test at n=k+1

    End of attempt

    To finish this step confuses me. I implore the forum for assistance.
    Last edited by Jameson; November 2nd 2009 at 09:01 AM. Reason: fixed latex
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  2. #2
    MHF Contributor
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    Quote Originally Posted by I-Think View Post
    The sequence of positive numbers u_1, u_2, u_3,... is such that u_1<4
    and
     u_{n+1}=\frac{5u_n+4}{u_n+2}

    By considering 4-u_{n+1}, prove by induction that u_n<4 for all n\geq{1}

    My attempt N.B To make this easier to read I'm skipping all the official statements of the induction process
    4-u_{n+1}=\frac{4-u_n}{2+u_n}

    Let n=1
    4-u_2=\frac{4-u_1}{2+u_1}
    It is given that u_1<4 and positive so \frac{4-u_1}{2+u_1}<4
    u_2=4-\frac{4-u_1}{2+u_1}
    Hence u_2 is less than 4

    Assume true for n=k
    4-u_{k+1}=\frac{4-u_k}{2+u_k}

    Test at n=k+1

    End of attempt

    To finish this step confuses me. I implore the forum for assistance.
    You're so close!

    You have a nice way of writing 4-u_{k+1} in terms of u_k. So now you need to show that \frac{4-u_k}{2+u_k}<4, which in turn shows that 4-u_{k+1}<4. This should be easy for you, since you have assumed that u_k < 4. Use that to show that the above fraction is < 4, thus 4-u_{k+1}<4, thus u_{k+1}<4. This shows that u_{k} < 4 implies u_{k+1}<4, which is the proof.
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