Find limit $ \lim_{n \to \infty}\left(1-\frac{1}{n}\right)\left(1 - \frac{2}{n}\right) \dots \left(1 - \frac{ \lfloor xn \rfloor -1}{n}\right). $

Question

Let $x >0$. Evaluate the limit: $$ \lim_{n \to \infty}\left(1-\frac{1}{n}\right)\left(1 - \frac{2}{n}\right) \dots \left(1 - \frac{ \lfloor xn \rfloor -1}{n}\right). $$

I am not sure how to show this since $(1-1/n)(1-2/n)$ seem to approach $1$ while $1-(\lfloor xn \rfloor-1)/n$ seems too approach $x$. Does the limit somewhat involve $e$ or even exist?

Answer 1

Hint:

Let $m:=\lfloor nx\rfloor$. The product is

$$\frac{(n-1)!}{(n-m)!n^{m-1}}.$$

By Stirling,

$$\sim\frac{\sqrt{n-1}\left(\dfrac{n-1}e\right)^{n-1}}{\sqrt{n-m}\left(\dfrac{n-m}e\right)^{n-m}n^{m-1}} \\\sim\frac{\sqrt{n-1}\left(\dfrac{n-1}n\right)^{n-1}e^{m-1}}{\sqrt{n}\sqrt{1-x}(1-x)^{n-m}} \\\sim\frac1{e^2\sqrt{1-x}}\left(\frac{e^x}{(1-x)^{1-x}}\right)^n.$$

Answer 2

• For $\ x=1\ $ the product is $\ \displaystyle\frac{(n-1)!}{n^{n-1}}= \frac{n!}{n^n}\ $. Now $\ \displaystyle\lim_{n\rightarrow\infty}\frac{(n!)^\frac{1}{n}}{n}=\frac{1}{e}\ $, so $\ \displaystyle\lim_{n\rightarrow\infty} \frac{n!}{n^n}=0\ $
• For $\ 1<x\ $, and sufficiently large $\ n\ $, $\ \lfloor xn\rfloor\ge n+1\ $ and the product will include the factor $\ \left(1-\frac{n+1-1}{n}\right)=0\ $. So in this case too, $\ \displaystyle \lim_{n\rightarrow\infty}\prod_{i=1}^{\lfloor xn\rfloor-1}\left(1-\frac{i}{n}\right)=0\ $.

• If $\ 0<x\le1\ $, let $\ 0<\alpha<\frac{x}{2(1-x)}\ $ and $\ m\ $ be any positive integer greater than $\ \frac{ \alpha x(1-x)}{x-2\alpha(1-x)}\ $. Then $\ 1<\frac{m(x-2\alpha(1-x))}{\alpha x(1-x)}=$$\left(\frac{m}{\alpha(1-x)}-\frac{2m}{x}\right)\ $. Thus the interval $\ \left(\frac{2m}{x}, \frac{m}{\alpha(1-x)}\right)\ $ must contain at least one integer. Therefore let $\ n\ $ be such an integer: $\ \frac{2m}{x}<n< \frac{m}{\alpha(1-x)}\ $.

  Now for $\ i\ge m\ $, $\ 1-\frac{i}{n}\le$$1-\frac{m}{n}<1-\alpha(1-x)\ $, so \begin{align} \prod_{i=1}^{\lfloor nx\rfloor-1} \left(1-\frac{i}{n}\right)&\le\prod_{i=m}^{2m-1} \left(1-\frac{i}{n}\right)\\ &<(1-\alpha(1-x))^{m-1}\ . \end{align} Since $\ 0< 1-\alpha(1-x)<1\ $, and $\ m\ $ can be arbitrarily large, it follows that $\ \displaystyle \lim_{n\rightarrow\infty}\prod_{i=1}^{\lfloor xn\rfloor-1}\left(1-\frac{i}{n}\right)=0\ $ in this case also.