$$\lim_{n \to \infty} \frac{\sum_{ k= 0}^{n-1} \prod_{i=0}^k (1-\frac{i}{n}) }{\sqrt{n}}$$ This problem arises from a probability question, which was inappropriately approximated into this form. After conducting computer simulations, the approximate result for this problem is around 1.25289. My classmate suggested that this can be viewed as randomly selecting one box out of n boxes without replacement until an empty box is chosen. They claimed that the expected number of balls can be transformed into an integral, possibly something like $$\int_{0}^{\infty} e^{-\frac{t^2}{2}} dt $$, which resembles $\sqrt{\frac{\pi}{2}}$. The computer simulations yielded a similar result. I would like to know if this is correct or if there is a more rigorous approach.
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n is an integer, I forgot to include it in the conditions. – math Jun 07 '23 at 04:05
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If you are familiar with Pochhammer symbols and their equivalents in terms of the gamma function $$\prod_{i=0}^k \left(1-\frac{i}{n}\right)=(-1)^k\,\frac{(1-n)_k }{n^k}$$ $$S_n=\sum_{k=0}^{n-1}\prod_{i=0}^k \left(1-\frac{i}{n}\right)=\sum_{k=0}^{n-1}(-1)^k\,\frac{(1-n)_k }{n^k}$$ $$S_n=e^n \,n^{1-n}\, \Gamma (n,n)$$
Looking here, @Ron Gordon gave $$\Gamma(n,n) = n^n e^{-n} \left [ \sqrt{\frac{\pi}{2 n}} - \frac{1}{3 n} + O\left ( n^{-3/2} \right ) \right ]$$
Using it $$S_n=\sqrt{\frac{\pi }{2}n} -\frac{1}{3}+ O\left ( n^{-1/2} \right )$$ which is a superb approximation even for small values of $n$ (relative error smaller than $1.0$% for $n>8$, smaller than $0.1$% for $n>83$.
Then, the answer.
Claude Leibovici
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