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Exponential inequality

  1. Oct 11, 2007 #1
    1. The problem statement, all variables and given/known data

    Prove: n! >or= (n^n)e^(1-n)

    Edit: For some positive integers n. (I'm very sure its all)


    3. The attempt at a solution

    Proof by Induction:

    Base: n=1
    1 >or= (1^1)e^0 = 1

    Induction Step (n->n+1)

    (n+1)! = (n+1)n! >= (n+1)(n^n)e^(1-n) = (n+1)(n^n)ee^(-n)

    Now I want to show (n+1)(n^n)e is greater or equal to (n+1)^(n+1)

    Taking natural log of both sides we get:

    ln(n+1) + nln(n) + 1 >=? (n+1)ln(n+1)

    nln(n) + 1 >=? nln(n+1)

    ne^(1/n) >=? n+1

    e >=? (1+1/n)^n

    Here at the very end of my proof is where I get stuck. I know that e can be defined as the limit as n approachs infinity of (1+1/n)^n but I'm unsure how to show that its greater then the sequence at all specific values of n. Any help would be appreciated.
     
    Last edited: Oct 12, 2007
  2. jcsd
  3. Oct 11, 2007 #2

    Dick

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    Ok, so you want to show (1+1/n)^n not only approaches e, but that it is increasing. You could do this by showing (1+1/x)^x is an increasing function for x>=1. I.e. show it's derivative is positive. The only way I've figured out so far is looking at power series expansions of log(1+x). It's not very elegant. Maybe you can think of something better.
     
  4. Oct 12, 2007 #3
    Alright I think I solved the last part.

    e >=? (1+1/n)^n

    then

    [ne^(1/n) - 1]/n >=? 0

    which occurs iff

    ne^(1/n)-1 >=? 0

    e^(1/n) >=? 1/n

    1/n >=? ln(1/n)

    Since 1>= ln(1) and ln(1/n) becomes negative for higher n it holds that 1/n>=ln(1/n) and therefore e >= (1+1/n)^n. Thus our induction is finished.

    Edit: I'll rewrite the proof for class to be more presentable, but I was wondering if this is totally rigorous?
     
    Last edited: Oct 12, 2007
  5. Oct 12, 2007 #4

    Gib Z

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    I have a nice solution, but it uses a more powerful result than your question :(
     
  6. Oct 12, 2007 #5

    Dick

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    It's worse than not totally rigorous. It's wrong. The second line "[ne^(1/n) - 1]/n >=? 0" should be >=1.
     
  7. Oct 12, 2007 #6

    Dick

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    Can you use the power series expansion of e^x at x=0? That's a nice way to show e^(1/n)>(1+1/n).
     
  8. Oct 12, 2007 #7
    Alright, I fixed my mistake:

    e >=? (1+1/n)^n

    e^(1/n) >=? 1 + 1/n

    n(e^(1/n)-1) >=? 1

    But e^x >=? 1+x since if you take the derivative of either side:

    e^x >= 1 Thus e^x has the the same tangent line at x=0 and a strictly greater slope for x > 0 and since e^x = x+1 at x=0 we know that e^x >= x+1 for x>0 (True for x<0 but not needed here).

    Thus n(e^(1/n)-1) >= n(1/n + 1 - 1) = 1 >= 1. Thus the inequality holds.

    BTW: Thanks for catching my mistake, I wouldnt have caught it in all likelihood.
     
  9. Oct 12, 2007 #8

    Dick

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    I think that's it.
     
  10. Oct 13, 2007 #9

    Gib Z

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    If you don't mind using the strong result, the inequality is trivial with Stirling's series.
     
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