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Consider $$ X_k =\prod_{\ell=1}^{k-2}\frac{n-\ell}{n} \ \textrm{ for every } 2\leqslant k\leqslant n+1. $$ How can you prove the following? $$ \lim_{n\rightarrow \infty} \frac1{\sqrt{n}}\sum_{k=2}^{n+1} X_k= \sqrt{\frac{\pi}{2}} $$ A heuristic argument replaces $\frac{n-\ell}{n}$ by $e^{-\ell}$ and the sum by an integral. How can this (or another method) be made rigorous? |
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Using the change of variable $k=n+1-i$ and some simple algebraic manipulations, one sees that the $n$th sum $S_n$ is $$ S_n=\frac1{\sqrt{n}}\frac{n!}{n^n}\sum_{i=0}^{n-1}\frac{n^i}{i!}=\frac1{\sqrt{n}}\frac{n!}{n^n}\mathrm e^n\mathbb P(Z_n\leqslant n-1), $$ where $Z_n$ is a random variable with Poisson distribution of parameter $n$. Thus, $Z_n$ is distributed as $Y_1+\cdots+Y_n$, where the sequence $(Y_n)_{n\geqslant1}$ is i.i.d. with Poisson distribution of parameter $1$. The central limit theorem applied to the sequence $(Y_n)_{n\geqslant1}$ shows that $T_n=(Z_n-n)/\sqrt{n}$ is asymptotically standard normal when $n\to\infty$, and in particular, $$ \mathbb P(Z_n\leqslant n-1)=1-\mathbb P(T_n\geqslant0)\to1-\tfrac12=\tfrac12. $$ This asymptotics, together with Stirling's approximation formula $n!\sim\sqrt{2\pi n}(n/\mathrm e)^n$, implies that $S_n\to\sqrt{\pi/2}$ when $n\to\infty$. |
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