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Thread: Proof by induction

  1. #1
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    Proof by induction

    Let a1=3 and a(n+1) =($\displaystyle {a}^{2}n+10$)/7 for every n$\displaystyle \geq$1

    Prove by induction that 2
    $\displaystyle \leq$an$\displaystyle \leq$5 for every n$\displaystyle \geq$1

    Deduce that an+1$\displaystyle \leq$an for every n$\displaystyle \geq$1

    Show the sequence {an} converges and determine its limit.
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  2. #2
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    Quote Originally Posted by Mathman87 View Post
    Let a1=3 and a(n+1) =($\displaystyle {a}^{2}n+10$)/7 for every n$\displaystyle \geq$1

    Prove by induction that 2$\displaystyle \leq$an$\displaystyle \leq$5 for every n$\displaystyle \geq$1

    Deduce that an+1$\displaystyle \leq$an for every n$\displaystyle \geq$1

    Show the sequence {an} converges and determine its limit.

    Induction on $\displaystyle n $ , $\displaystyle a_1 =3 \in[2,5]$ and assume it is true that $\displaystyle 2 \leq a_n \leq 5$ and we have

    $\displaystyle a_{n+1} = \frac{a_n^2 + 10}{7} \leq \frac{ 5^2 + 10}{7} = \frac{35}{7} = 5$

    Also ,

    $\displaystyle a_{n+1} = \frac{a_n^2 + 10}{7} \geq \frac{2^2 + 10}{7} = \frac{14}{7} = 2 $

    Therefore , $\displaystyle 2\leq a_{n+1} \leq 5 $

    Hence we deduce it is true that $\displaystyle 2\leq a_n \leq 5 ~~\forall n \in \mathbb{N} $

    Consider $\displaystyle a_{n+1} - a_n = \frac{a_n^2 + 10}{7} -a_n = \frac{a_n^2 - 7a_n + 10 }{7} = \frac{(a_n - 2)(a_n -5 )}{7}$

    $\displaystyle a_n - 2 \geq 0 $ because $\displaystyle a_n \geq 2$

    but $\displaystyle a_n - 5 \leq 0 $ from $\displaystyle a_n \leq 5 $

    Therefore , $\displaystyle a_{n+1} - a_n$ is a non-positive number or $\displaystyle a_{n+1} - a_n \leq 0 ~,~ a_{n+1} \leq a_n $

    Note that the sequence is bounded and it is monotonic decreasing , use some theorems to show that the limit exists and find it .
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