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Math Help - Proof by induction and sequences

  1. #1
    Senior Member chella182's Avatar
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    Proof by induction and sequences

    The question goes...

    Let the sequences \{a_{n}\} and \{b_{n}\} be defined by a_{1}=1, b_{1}=1,
    a_{n+1}=a_{n}+\frac{1}{n(n+1)} and b_{n+1}=b_{n}+\frac{1}{(n+1)^2}.
    Prove by induction that a_{n}=2-\frac{1}{n} and b_{n}\leq a_{n}. What is \lim_{n\rightarrow\infty}a_{n}? Prove that \lim_{n\rightarrow\infty}b_{n} exists.

    Can I have quite a detailed explanation please? I was never any good at proof by induction & I'm not too good at pure maths in general. Plus my lecturer's not been available for meetings lately
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  2. #2
    Junior Member Infophile's Avatar
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    Hello,

    Initialisation : a_1=2-\frac{1}{1}=1 true.

    Then suppose that a_n=2-\frac{1}{n}.

    a_{n+1}=a_n+\frac{1}{n(n+1)}=2-\frac{1}{n}+\frac{1}{n(n+1)}=2-\frac{(n+1)-1}{n(n+1)}=2-\frac{1}{n+1}

    What ends the proof by induction.

    Now, try to show by induction that \forall n\in \mathbb{N}, b_n\le b_n.

    Then it is clear that (b_n) increases and \forall n\in \mathbb{N}, b_n\le a_n\le 2 so (b_n) converge.

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  3. #3
    Senior Member chella182's Avatar
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    why suppose that a_{n}=2-\frac{1}{n}?
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  4. #4
    Junior Member Infophile's Avatar
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    It's the hypothesis of induction at rank n, and you show it stay true at rank n+1.
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  5. #5
    Senior Member chella182's Avatar
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    Okay. Still not 100% sure, but you're obviously right with your induction I still don't get the b_{n}\leq a_{n} bit though.
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  6. #6
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    Grandad's Avatar
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    Hello chella182
    Quote Originally Posted by chella182 View Post
    Okay. Still not 100% sure, but you're obviously right with your induction I still don't get the b_{n}\leq a_{n} bit though.
    To prove this, note first that for n>0, n(n+1)<(n+1)^2 \Rightarrow \frac{1}{n(n+1)}>\frac{1}{(n+1)^2}

    So for n>0, a_n \ge b_n \Rightarrow a_n + \frac{1}{n(n+1)}\ge b_n + \frac{1}{n(n+1)} > b_n + \frac{1}{(n+1)^2}

    \Rightarrow a_{n+1} > b_{n+1}

    And a_1 = b_1

    So by induction a_n \ge b_n,\, \forall n \ge 1

    Grandad
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  7. #7
    Senior Member chella182's Avatar
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    Thankyouu most helpful.
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