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Math Help - More sequence problems :(

  1. #1
    Junior Member simplysparklers's Avatar
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    More sequence problems :(

    Hiya guys!
    I'd really appreciate some help with this question, I don't even know where to begin!This book I'm using is rubbish!

    Consider the sequence (a_{n}), where
    a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3} (n\geq1)

    a.) Show by induction that 0\leq{a_{n}}<1

    b.) Show that a_{n} is monotonic increasing

    But where do I go about getting a_{n} in the first place?? I'd really appreciate a full explaination. Thanks so much guys!

    Jo
    Last edited by simplysparklers; May 16th 2008 at 12:06 PM. Reason: Latex problems again
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  2. #2
    Moo
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    Quote Originally Posted by simplysparklers View Post
    Hiya guys!
    I'd really appreciate some help with this question, I don't even know where to begin!This book I'm using is rubbish!

    Consider the sequence (a_{n}), where
    a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3} (n\geq1)

    a.) Show by induction that 0 \leq a_{n} <1

    b.) Show that a_{n} is monotonic increasing

    But where do I go about getting a_{n} in the first place?? I'd really appreciate a full explaination. Thanks so much guys!

    Jo
    Fixed it

    < or > is enough, you don't need any \
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  3. #3
    Junior Member simplysparklers's Avatar
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    Lol!Cool!Thanks Moo!!I'm still only getting the hang of it!!
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  4. #4
    Moo
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    For question a :

    Basis : 0 \leq a_1=0 < 1

    Induction hypothesis : 0 \leq a_n <1

    (--> is 0 \leq a_{n+1} <1 ?)


    According to the hypothesis, 3 \leq a_n+3<4

    --> \frac 13 \geq {\color{blue}\frac{1}{a_n+3}} > \frac 14 \Longleftrightarrow \frac 14 < {\color{blue}\frac{1}{a_n+3}} \leq \frac 13 (for more convenience )

    Also according to the hypothesis, 0 \leq 3a_n <3
    1 \leq {\color{red}3a_n+1} < 4


    --->  ?? < \underbrace{({\color{red}3a_n+1}) \cdot  {\color{blue}\frac{1}{a_n+3}}}_{a_{n+1}} < ??

    Can you continue ?



    Quote Originally Posted by simplysparklers View Post
    Lol!Cool!Thanks Moo!!I'm still only getting the hang of it!!
    No problem
    You'll get used to it by practicing ^^









    Edit : Ok, I didn't check it at first time, but my last inequality doesn't yield anything interesting
    Last edited by Moo; May 16th 2008 at 12:28 PM. Reason: parenthesis
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  5. #5
    Junior Member simplysparklers's Avatar
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    Quote Originally Posted by Moo View Post
    --->  ?? < \underbrace{({\color{red}3a_n+1}) \cdot  {\color{blue}\frac{1}{a_n+3}}}_{a_{n+1}} < ??

    Can you continue?









    Edit : Ok, I didn't check it at first time, but my last inequality doesn't yield anything interesting

    Hey Moo!
    Are you saying then that this doesn't actually prove that 0\leq{a_n}<1??
    Because this:  ?? < \underbrace{({\color{red}3a_n+1}) \cdot  {\color{blue}\frac{1}{a_n+3}}}_{a_{n+1}} < ??
    is actually:
    0\leq{a_{N+1}}<1, isn't it??

    And any ideas on part b then anyone??Thanks for your help!!
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  6. #6
    Moo
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    Quote Originally Posted by simplysparklers View Post
    Hey Moo!
    Are you saying then that this doesn't actually prove that 0\leq{a_n}<1??
    Because this:  ?? < \underbrace{({\color{red}3a_n+1}) \cdot  {\color{blue}\frac{1}{a_n+3}}}_{a_{n+1}} < ??
    is actually:
    0\leq{a_{N+1}}<1, isn't it??

    And any ideas on part b then anyone??Thanks for your help!!
    Actually, I wanted you to fill in the "??" gaps, with the previous inequalities, but it doesn't yield 0 and 1 ~


    For b), show that a_{n+1}>a_n, this has to do with a quadratic inequation
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  7. #7
    Junior Member simplysparklers's Avatar
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    So is the answer to the first one: \frac{1}{4}\leq{(3a_{n}+1).\frac{1}{a_{n}+3}}<\fra  c{4}{3}
    ..which holds true, therefore, by induction, 0\leq{a_{n}}<1 fpr all n?

    And could you elaborate on part b please?I get that (a_{n}) is monotonic increasing if a_{n+1}\geq{a_{n}} for all n\geq{1}, but how do you prove this?When you don't know what (a_{n}) is??
    Coz like that, the last part of this questionis: Find \lim_{n\rightarrow\infty}a_{n}. I think I could do this part if I just knew where to start with getting a_{n}
    I know I'm probably being really stupid about this, sorry!
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  8. #8
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    Quote Originally Posted by simplysparklers View Post
    Hiya guys!
    I'd really appreciate some help with this question, I don't even know where to begin!This book I'm using is rubbish!

    Consider the sequence (a_{n}), where
    a_{1}=0, a_{n+1}=\frac{3a_{n}+1}{a_{n}+3} (n\geq1)

    a.) Show by induction that 0\leq{a_{n}}<1

    b.) Show that a_{n} is monotonic increasing

    But where do I go about getting a_{n} in the first place?? I'd really appreciate a full explaination. Thanks so much guys!

    Jo
    Proving a_n \geq 0 by induction is almost trivial.

    For a_n < 1,by induction, do the following:

    a_{n+1} = 3 - \frac{8}{a_{n}+3}

    a_n < 1 \Leftrightarrow a_{n}+3 < 4 \Leftrightarrow -\frac1{a_{n}+3} < -\frac14 \Leftrightarrow a_{n+1} = 3-\frac8{a_{n}+3} < 1
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  9. #9
    Junior Member simplysparklers's Avatar
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    Thank you Isomorphism! I get the final line, but to begin with, where did you get a_{n+1}=3-\frac{8}{a_{n}+3} in the first place??Is it really obvious and I'm just not seeing it?? My brain seems to work extra slow on sequences
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  10. #10
    MHF Contributor Mathstud28's Avatar
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    Quote Originally Posted by simplysparklers View Post
    Thank you Isomorphism! I get the final line, but to begin with, where did you get a_{n+1}=3-\frac{8}{a_{n}+3} in the first place??Is it really obvious and I'm just not seeing it?? My brain seems to work extra slow on sequences
    \frac{3a_n+1}{a_n+3}=\frac{3a_n+9-8}{a_n+3}=\frac{3a_n+9}{3a_n+3}-\frac{8}{a_n+3}=\frac{3(a_n+3)}{a_n+3}-\frac{8}{a_n+3}=3-\frac{8}{a_n+3}
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