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So how to approximate $c_3=\ {\infty +a-1 \choose \infty-1}$

$c_2=\ {\infty \choose \infty-2-a+b}$

$c_1=\ {\infty +a-b +1 \choose \infty-1}$

$p=\ {(c_1-c_2)/c_3}$

I'm trying to get the limit of this cumulative probability. It's like stacking 'a' balls in infinite boxes and finding the probability that the stack is at least 'b' high

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I don't understand what the factorial of $\infty$ is supposed to mean, but why not apply Stirling's formula (c.f. en.wikipedia.org/wiki/Stirling's_approximation) to the definition of the binomial symbol? This should give you a finite limit to the above problems as you approach $\infty$. –  A Blumenthal Feb 5 '13 at 19:38
    
I think this question is equivalent with this one $p=\lim_{n\to\infty}\frac {c_1-c_2}{c_3}=\lim_{n\to\infty}\frac {\binom{n+a-b +1}{n-1}-\binom{n}{n-2-a+b}}{\binom {n+a-1}{n-1}}$ –  Adi Dani Feb 5 '13 at 19:40
    
applying stirlings formula a lot of tedious work –  Jennifer Aneta Feb 5 '13 at 19:51

1 Answer 1

For every $n$ large enough and for $a\gt b$ positive, define $$ p(n)=\frac{{n +a-b +1 \choose n-1}-{n \choose n-2-a+b}}{{n +a-1 \choose n-1}}. $$ Then, when $n\to\infty$, $$ p(n)=\frac{a!}{(a-b)!}n^{1-b}+o(n^{1-b}). $$ Is this the kind of result you are after?

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