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Proving inequality by mathematical induction

  1. May 9, 2012 #1
    1. The problem statement, all variables and given/known data

    I am asked to prove:
    2n < (n+1)! , where n≥2

    3. The attempt at a solution

    Base step: set n=2, then test 22 < (2+1)!

    22 = 4
    (2+1)!= 3! = 3(2)(1) = 6
    so 4 < 6 , which is true.

    Induction hypothesis is 2k < (k+1)!
    Using this, prove 2(k+1) < [(k+1)+1]! = (k+2)!

    Attempt to solve:

    starting with what I know: 2k < (k+1)!
    Multiplying both sides by 2: 2(2k) = 2(k+1) < 2(k+1)!

    I know that 2(k+1)! < (k+2)!
    since (k+2)! = (k+2)(k+1)! and because k≥2, (k+2) will be greater than 2. Thus, multiplying (k+1)! by 2 on the LHS is less than multiplying (k+1)! by (k+2) on the RHS.

    Thus, since 2(k+1) < 2(k+1)! is true, then 2k+1 < [(k+1)+1]!.

    P(k+1) follows from P(k), completing the induction step. By mathematical induction, P(n) is true for n≥2.


    Thanks for any help!
    EDIT: fixed a couple of type-o's.
     
    Last edited: May 9, 2012
  2. jcsd
  3. May 9, 2012 #2

    Mark44

    Staff: Mentor

    Welcome to PF, and very nice first post! Looks good!
     
  4. May 9, 2012 #3
    Thank you very much for your response! It was posted as an extra credit problem by my professor. I wanted to make sure my reasoning was correct before posting it on the board, as I've been having a little difficulty grasping what we covered about mathematical induction.
     
  5. May 9, 2012 #4

    Mark44

    Staff: Mentor

    It looks to me like you have the idea of induction proofs.
     
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