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A difficult limit for me

  1. Jan 6, 2009 #1
    1. The problem statement, all variables and given/known data
    show that [tex]\displaystyle \lim_{n \to \infty} \left[ \left(\begin{matrix} n \\ 0 \end{matrix} \right) \left(\begin{matrix} n \\ 1\end {matrix} \right) ....\left(\begin{matrix} n \\ n \end{matrix} \right ) \right]^\frac{1}{n^2} = e^\frac{1}{2}[/tex]

    2. Relevant equations

    Stirling's approximation [tex]n! \sim \sqrt{2 \pi n} n^n e^{-n}[/tex]

    3. The attempt at a solution
    Firstly I tried to use smallest term to the n-power because that is # of terms of these combinations. Then, [tex] \displaystyle \lim_{n \to \infty} ((n^n))^\frac{1}{n^2} =1 [/tex]. Secondly I did the same with de largest term, taking into account that middle term of Pascal's triangle [tex]\displaystyle \sim \frac{n!}{(\frac{n}{2}!)(\frac{n}{2}!)}[/tex] which gave me 2, so now I know that the limit is between 2 and 1 but I haven't proved that the limit is [tex]e^\frac{1}{2}[/tex]. I also tried logs and play with these combinations from right to left and in reverse but it did not help much. If anybody knows any generating function or idea that I can apply, it will be greatly appreciated.
  2. jcsd
  3. Jan 7, 2009 #2


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    You can write the product of the C(n,k) as (n!)^n/(0!*1!*...*n!)^2. Take the log and apply Stirling's approximation log(n!)~n*log(n)-n. To estimate the sum of k*log(k) from 1 to n, approximate it by the integral of x*log(x) (much the same way you derive a rough version of Stirling's approximation).
    Last edited: Jan 7, 2009
  4. Jan 7, 2009 #3
    thank you.

    After your advice everything was really easy. I didn't look elegant, however, it was effective.
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