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Inequality proof

  1. Oct 15, 2014 #1
    1. The problem statement, all variables and given/known data


    mathshity.JPG

    its the second one.
    let n∈ℕ \ 0 and k∈ℕ show that
    (n choose k) 1/n^k <= 1/k!

    2. Relevant equations

    axioms of ordered fields?




    3. The attempt at a solution

    i have been working on this all afternoon. I know 0<k<n since its a requirement for (n choose k). Ive tried induction, base case works but cannot figure out how to do inductive step. I dont really think induction is the correct method. Ive tried cases, n=k is easy to prove but n>k i cannot figure out. Any help getting me started in the right direction would be appreciated.
     
  2. jcsd
  3. Oct 15, 2014 #2
    As n > 0 you can instead try proving the equivalent statement:

    [tex]\binom n k \le \frac {n^k}{k!}[/tex]
     
  4. Oct 15, 2014 #3

    RUber

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    ##n! < n^n ##
    So what inequality might hold for the largest k terms of n! ?
     
  5. Oct 15, 2014 #4
    Thanks for the replies.

    I tried playing around with these for an hour this morning. When trying to prove that equivalent statement, Its obvious that its true but I still don't know how to "prove" it. Leaves me in the same situation as before just allows me to cancel a factor of 1/k! from either side of the inequality.

    RUber, that inequality holds for all n>=2 i believe. Since k has to be less than n though the only way I could write k in terms of n is n-1 and at this point i am struggling again on what to do next.

    Right now I'm thinking cases is the best way to try and prove this since its easy to prove the inequality for n=k. I just need to prove that it holds for n>k and then I would be done.
     
  6. Oct 15, 2014 #5

    RUber

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    I don't think you need induction, this is a straightforward relation.
    ##\left( \begin{array}{c} n\\k \end{array} \right) \frac{1}{n^k} \leq \frac{1}{k!} ##
    Is saying:
    ##\frac{n!}{(n-k)!(k)!}\frac{1}{n^k} \leq \frac{1}{k!} ##
    Which is equivalent to (for k >0)
    ##\frac{\left( \prod_{i = n-k+1}^n i \right) }{(k)!}\frac{1}{n^k} \leq \frac{1}{k!} ##
    So as you said by cancelling out the ##\frac{1}{k!} ## the relation you need to prove is :
    ##\frac{\left( \prod_{i = n-k+1}^n i \right) }{n^k} \leq 1##
    And for k= 0 :
    ##\frac{n!}{(n-k)!(k)!}\frac{1}{n^k} =1 \leq \frac{1}{k!} =1 ##
     
  7. Oct 15, 2014 #6
    ok i'm able to understand most of this. Im a little confused how you changed n!/(n-k)! into that product but i can see that it makes sense. Isn't k=0 just one case though? Wouldn't i still need to prove somehow that the inequality holds for k>0 but still k<n?
     
  8. Oct 15, 2014 #7

    RUber

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    ## \frac{5!}{(5-2)!}= \frac{5*4*3*2*1}{3*2*1}=\prod_{i=5-2+1}^5 i =\prod_{i=4}^5 i=4*5.##
    The first relation I wrote above (with the product notation) is for any k not equal to zero, then for k=0, the relation is just 1=1.
    I find it odd that it seems your definition of natural numbers includes zero. That is not what I am used to. In any case, it just adds one case to the proof.
     
  9. Oct 16, 2014 #8
    Ok thanks im pretty sure I got it now.
     
  10. Oct 16, 2014 #9

    Ray Vickson

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    If you insist on posting images (which is discouraged in PF) you should at least attempt to post them correctly--not sideways like you did. If you want help, don't make it difficult for us to see what you are doing.
     
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