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Math Help - Trying to prove that 2^n < n! for n>= 4

  1. #1
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    Trying to prove that 2^n < n! for n>= 4

    Is my proof correct?

    2^n < n! for all n >= 4
    Step 1: solve for base case
    2^4 < 4! --> 16 < 24


    Assume p(n) is true i.e. 2^n < n! is true

    p(n+1)
    2^(n+1) < (n+1)!

    Try to solve through inductive reasoning.
    In this case, just multiply both sides by (n+1)
    2^n (n+1) < (n!)(n+1)

    Since
    n>1 we can say 2^n(n+1) > (2)2^n
    (2)2^n = 2^(n+1)

    So:
    (n+1)! = n!(n+1) > (2^n)(n+1) > (2)2^n = 2^(n+1)
    which means
    2^(n+1) < (n+1)!
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  2. #2
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    Quote Originally Posted by Bruins1 View Post
    Is my proof correct?

    2^n < n! for all n >= 4
    Step 1: solve for base case
    2^4 < 4! --> 16 < 24


    Assume p(n) is true i.e. 2^n < n! is true

    p(n+1)
    2^(n+1) < (n+1)!

    Try to solve through inductive reasoning.
    In this case, just multiply both sides by (n+1)
    2^n (n+1) < (n!)(n+1)

    Since
    n>1 we can say 2^n(n+1) > (2)2^n
    (2)2^n = 2^(n+1)

    So:
    (n+1)! = n!(n+1) > (2^n)(n+1) > (2)2^n = 2^(n+1)
    which means
    2^(n+1) < (n+1)!
    Yes ,i think so
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