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Yet another induction proof

  1. Dec 31, 2006 #1
    1. The problem statement, all variables and given/known data
    show with mathematical induction that prod(k=1->n) (1+1/n)^n=(n+1)^n/n!
    n element Natural numbers

    3. The attempt at a solution

    works for n=1,
    [tex]\prod_{k=1}^{n+1}\left(1+\frac{1}{k+1}\right)^{k+1}[/tex] should become [tex]\frac{(n+2)^{n+1}}{(n+1)!}[/tex]
    [tex]= (1+(n+1)^{-1})*((1+(n+1)^{-1})^n)*\frac{(n+1)^n}{n!}[/tex]
    [tex]=\frac{(n+1)^n}{n!})*(1+(n+1)^{-1})(1+(n+1)^{-n})[/tex]
    [tex]=\frac{(n+1)^n}{n!}*(1+(n+1)^n+(n+1)^{-1}+(n+1)^{-n+1}[/tex]
    [tex]=\frac{(n+1)^n}{n!}+\frac{1}{n!}+\frac{(n+1)^n}{(n+1)!}+\frac{(n+1)^n}{(n+1)^{n+1})n!}[/tex]
    [tex]=\frac{(n+1)^{n+1}+(n+1)+(n+1)^n+1)}{(n+1)!}[/tex]

    only i can't find any way to get to (n+2)^(n+1)/(n+1)! from there :(
    can anyone help?
    thanks in advance :)


    ps. Happy New Year to those it applies to:)
     
    Last edited: Jan 1, 2007
  2. jcsd
  3. Dec 31, 2006 #2

    radou

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    I might be missing something, but did you perhaps mean
    [tex]\prod_{k=1}^n \left(1+\frac{1}{k}\right)^k = \frac{(n+1)^n}{n!}[/tex]
    ?
     
  4. Dec 31, 2006 #3

    matt grime

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    I think you'd better take time out to learn latex if you're going to post things like that.

    Anyway, you'd better get your n's and k's sorted out. I can't see any immediate interpretation of what yo'uve written that might be true.
     
  5. Dec 31, 2006 #4
    yeah.. that'd be it.. i tried editing the original a bit to make it more clear what exactly i've tried.. hope i didn't include any errors
     
    Last edited: Dec 31, 2006
  6. Dec 31, 2006 #5

    radou

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    Ok, as you have seen, it holds for n = 1. Next, assume it holds for some n = j. You have to proove it holds for n = j + 1 in order to conclude it holds for every natural number. Now, write down what it looks like when n = j and when n = j + 1. The rest is easy.
     
  7. Dec 31, 2006 #6

    matt grime

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    It's easier to do without induction (just simplify the product), if you're allowed to. Doesn't sound like it.
     
  8. Dec 31, 2006 #7
    @radou: i know it's supposed to be easy, but the answer still eludes me
    @matt: i'm not :)
     
  9. Dec 31, 2006 #8

    radou

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    Well, you have
    [tex]\prod_{k=1}^{j+1}\left(1+\frac{1}{k}\right)^k = \prod_{k=1}^j \left(1+\frac{1}{k}\right)^k \cdot \left(1+\frac{1}{j+1}\right)^{j+1}[/tex].
    That should illustrate the point, I hope.
     
  10. Jan 1, 2007 #9
    well, sure.. but, in my (granted, rather limited) experience with these things you then substitute the righthand side of part one with the product bit, and multiply that out, which should yield the n+1 RH part of the equation.. which is what i believe i tried to do in my try as shown above.. which is where i get stuck.
     
  11. Jan 1, 2007 #10

    matt grime

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    Then try again. That is all you need to do: mutliply out

    [tex] \frac{(j+1)^{j}}{j!} \left(1 +\frac{1}{j+1}\right)^{j+1}[/tex]
     
    Last edited: Jan 1, 2007
  12. Jan 1, 2007 #11
    yes, but can you tell me if something has gone wrong in what i did in the OP? because it doesn't seem to add up to anything like the wanted result :S
     
    Last edited: Jan 1, 2007
  13. Jan 1, 2007 #12

    matt grime

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    I cannot even decipher the original post - there is too much going on and things seem to vanish between lines, then reappear, like the factorial.

    What is another way to write 1+ 1/(j+1)? It really is a one line simplification of the expression in my last post. If your original line of attack doesn't work try a different one.
     
  14. Jan 1, 2007 #13
    ah, yes.. i seem to have become somewhat obtuse after two years of not doing any maths :(

    thanks for your patience :)
     
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