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Math Help - Induction with inequality

  1. #1
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    Induction with inequality

    How do i use induction(effectively) to prove that n! < nn for n ≥ 2 ?

    let p(n): n! < nn for n ≥ 2.

    we see that p(2) holds true.

    now i prove it holds for p(n+1) right?

    (n+1)! = (n+1)n!<(n+1)nn by the induction hypothesis.


    i think this is right.. can someone show me how to finish this correctly?

    thanks

    Kyle
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  2. #2
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    oops

    oops, n^n came out as nn...anywhere there is nn, it should read (n)^n.
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  3. #3
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    Quote Originally Posted by CSkyle View Post
    How do i use induction(effectively) to prove that n! < nn for n ≥ 2 ?

    let p(n): n! < nn for n ≥ 2.

    we see that p(2) holds true.

    now i prove it holds for p(n+1) right?

    (n+1)! = (n+1)n!<(n+1)nn by the induction hypothesis.


    i think this is right.. can someone show me how to finish this correctly?

    thanks

    Kyle

    assume k! < k^k

    Show

    (k+1)^{k+1}=(k+1)^k(k+1) > k^k(k+1)>(k!)(k+1)=(k+1)!

    So

    (k+1)^{k+1}>(k+1)!

    QED
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  4. #4
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    thx

    thanks empty set.
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