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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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looks like some tricky application of CLT may do the job –  K.Ghosh Jan 1 '13 at 11:32
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up vote 5 down vote accepted

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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"This" is "universally true" (as you say) as long as "this" corresponds to the precise and rigorous definition of asymptotic equivalence which is meant in my post (the trouble here being that the symbol $\sim$ is sometimes used with a less precise meaning, in which case, as you say, more caution would be needed). –  Did Jan 1 '13 at 17:18
    
Except that if $f(x)=1/x$ and $g(x)=0$, then $g(x)\sim f(x)$ is quite wrong. (Exercise: assume that $g(x)=0$ for every $x$ and that $g(x)\sim k(x)$ when $x\to x_0$, then $k(x)=0$ for every $x$ in a neighbourhood of $x_0$.) –  Did Jan 1 '13 at 17:49
    
...anything but itself, that is. –  Did Jan 1 '13 at 17:55
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Not. At. All. The whole point of asymptotics analysis is to compare bunches of functions with the same limit. Thus one wants to say, for example, that $x\mapsto x$ and $x\mapsto\sin(x)$ are roughly the same when $x\to0$ but that $x\mapsto x$ and $x\mapsto x^2$ are not. –  Did Jan 1 '13 at 18:06
    
Point taken. Thanks. –  user54551 Jan 1 '13 at 18:14
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