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Is this weird limit true?

  1. Dec 21, 2011 #1
    Here's the weird limit:

    lim n[itex]\rightarrow[/itex] [itex]\infty[/itex] aCn/na = 1/(a!)

    Don't ask how I thought this up, but let me explain my reasoning.
    Let's say that a=3. Then aCn = (n)(n-1)(n-2)
    And because n is approaching infinity, could it be (n)(n)(n) = n3?

    I wouldn't normally think of things like this but if it's true then it helps me out.

    The only thing is that (n)(n-1)(n-2)= n3 +something*n2+something*n

    And when I graphed it it looked like it approaches zero.
    I've never learned limits formally. Thanks.

    EDIT: aCn is a Choose n, or n!/(a![n-a]!)
    EDIT: Wolfram alpha is amazing http://www.wolframalpha.com/input/?i=lim+(x!/(3!(x-3)!x^3))+as+x->+infinity
    Sorry for not trying this before posting. I'd still be interested in other replies as to why this is true, though
     
    Last edited: Dec 21, 2011
  2. jcsd
  3. Dec 22, 2011 #2

    HallsofIvy

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    No, because n is not fixed, it is, as you said, approaching infinity.
    In any case, aCn= n(n-1)(n-3)/3!, not just n(n-1)(n-2). That goes to infinity as n goes to infinity.

    Now, this is completely different- you have an [itex]n^3[/itex] in the denominator.
    [tex]\frac{_nC_3}{n^3}= \frac{n!}{3!(n-3)!}\frac{1}{n^3}= \frac{n(n-1)(n-2)}{6n^3}[/tex]
    Since both numerator and denominator are of third power in n, that limit is 1/6.
     
    Last edited: Dec 22, 2011
  4. Dec 22, 2011 #3
    I have not tried yet, but a stirling's approximation of [itex]{}_aC_n[/itex] would give the correct limit
    [tex]n!\sim\sqrt{2\pi n}\left(\frac{n}{e}\right)^n[/tex]
     
  5. Dec 24, 2011 #4
    No, you're limit is correct. It's also not that weird. Consider the expansion of [itex]{n\choose a}=\frac{n(n-1)\ldots (n-a+1)}{a!}[/itex]. Then, consider that [itex]\frac{n(n-1)\ldots (n-a+1)}{n^a}=1\cdot (1-\frac{1}{n})\cdot\ldots\cdot (1-\frac{a-1}{n})[/itex]. Each of those factors goes to 1 as [itex]n\rightarrow\infty[/itex], thus the whole product goes to 1 as well. The expression [itex]{n\choose a}n^{-a}[/itex] is the same thing, but divided by [itex]a![/itex].
     
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