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Math Help - Proof by induction

  1. #1
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    Proof by induction

    Let a1=3 and a(n+1) =( {a}^{2}n+10)/7 for every n \geq1

    Prove by induction that 2
    \leqan \leq5 for every n \geq1

    Deduce that an+1 \leqan for every n \geq1

    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) =( {a}^{2}n+10)/7 for every n \geq1

    Prove by induction that 2 \leqan \leq5 for every n \geq1

    Deduce that an+1 \leqan for every n \geq1

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

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

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

    Also ,

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

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

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

    Consider  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}

     a_n - 2 \geq 0 because  a_n \geq 2

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

    Therefore ,  a_{n+1} - a_n is a non-positive number or  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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