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Thread: 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 $\displaystyle \{a_{n}\}$ and $\displaystyle \{b_{n}\}$ be defined by $\displaystyle a_{1}=1, b_{1}=1$,
    $\displaystyle a_{n+1}=a_{n}+\frac{1}{n(n+1)}$ and $\displaystyle b_{n+1}=b_{n}+\frac{1}{(n+1)^2}$.
    Prove by induction that $\displaystyle a_{n}=2-\frac{1}{n}$ and $\displaystyle b_{n}\leq a_{n}$. What is $\displaystyle \lim_{n\rightarrow\infty}a_{n}$? Prove that $\displaystyle \lim_{n\rightarrow\infty}b_{n}$ 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 : $\displaystyle a_1=2-\frac{1}{1}=1$ true.

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

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

    What ends the proof by induction.

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

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

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  3. #3
    Senior Member chella182's Avatar
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    why suppose that $\displaystyle a_{n}=2-\frac{1}{n}$?
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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 $\displaystyle b_{n}\leq a_{n}$ 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 $\displaystyle b_{n}\leq a_{n}$ bit though.
    To prove this, note first that for $\displaystyle n>0, n(n+1)<(n+1)^2 \Rightarrow \frac{1}{n(n+1)}>\frac{1}{(n+1)^2}$

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

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

    And $\displaystyle a_1 = b_1$

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

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