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Homework Help: Given as true f(x) = (1+1/x)^x is strictly increasing for x>=1

  1. Sep 19, 2010 #1
    1. The problem statement, all variables and given/known data
    Given as true f(x) = (1+1/x)^x is strictly increasing for x>=1 and that f(x) has horizontal asymptote y=e.

    Prove that (n/3)^n< n! <(n/2)^n for all integers n>=6 ?

    2. Relevant equations



    3. The attempt at a solution
    f(x)=(1+1/x)^x is increasing and approach e
    prove (n/3)^n< n! <(n/2)^n for all n>=6

    So I attempt to use f(x) to replace n but that will not work for the base case 1 because n>6
     
    Last edited: Sep 19, 2010
  2. jcsd
  3. Sep 19, 2010 #2

    Mark44

    Staff: Mentor

    Re: Induction

    The base case doesn't have to be n = 1. Use n = 6 for your base case.
     
  4. Sep 19, 2010 #3
    Re: Induction

    Thanks Mark44

    So I use k=F(x) from first function

    Base k=6;

    (k/3)^k< k! <(k/2)^k

    64<180<729 true

    K+1->;
    ((k+1)/3)^(k+1)< (k+1)! <((k+1)/2)^(k+1)

    But why was I given as true f(x) = (1+1/x)^x all I did was make K=f(x)
    Do I have to work out how to get a 6 from that function.( I don't think its possible)
     
    Last edited: Sep 19, 2010
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