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Proof by induction: n^3 < n! for n >=6

  1. Mar 8, 2015 #1
    1. The problem statement, all variables and given/known data
    Show that n^3 < n! for all n >= 6.

    2. Relevant equations


    3. The attempt at a solution

    We see that for the base case of n = 6, the claim holds.

    Suppose that k^3 < k! for some natural number k >= 6.

    Consider that:
    (k+1)^3
    = k^3 + 3k^2 + 3k + 1
    < k! + 3k^2 + 3k + 1 [By induction hypothesis]

    What's a neat way to finish this? I'm a bit rusty, apparently.
     
  2. jcsd
  3. Mar 8, 2015 #2

    Dick

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    Factor a ##k^3## out of ##k^3+3k^2+3k+1##. Then use ##k^3 \le k!## and ##(k+1)!=k!(k+1)##.
     
  4. Mar 9, 2015 #3
    Alright. And then, what remains after being factored has its largest value at k=6, and its value is smaller than any (k+1), and so I may write < k!(k+1), completing the induction.
     
  5. Mar 9, 2015 #4

    Dick

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    Sounds ok.
     
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