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Proof by induction.

  1. Jul 14, 2011 #1
    1. The problem statement, all variables and given/known data
    Let n be a natural number.

    Prove that 0!+1!+2!+.......n! < (n+1)!
    and my < sign should be less than or equal to.
    3. The attempt at a solution

    The proof is by induction
    it works for 0 because 0! is less than or equal to 1!
    now we assume it works for n=k by the induction axiom.
    now we see if it works for k+1
    0!+1!+2!+.....k!+(k+1)!<(k+1+1)!
    Now im not sure if I can do this but I will replace
    0!+1!+2!+.....k! with (k+1)!
    so now I have (k+1)!+(k+1)!<(k+2)!
    2(k+1)!<(k+2)!
    2(k+1)!<(k+2)(k+1)!
    and k+2 will always be bigger or equal to 2 because k is a natural number.
    so the inequality is proved by induction.
     
  2. jcsd
  3. Jul 15, 2011 #2

    Dick

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    Science Advisor
    Homework Helper

    That works fine. Your induction hypothesis is 0!+1!+2!+.......k! <= (k+1)!. So 0!+1!+2!+.....k!+(k+1)! <= (k+1)!+(k+1)!.
     
  4. Jul 16, 2011 #3
    ok thanks so everything i did is ok.
     
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