Asymptotic behaviour of infinite product of ceilings $\lceil\alpha\rceil \lceil2\alpha\rceil \lceil3\alpha\rceil\dots$, $\alpha\in(0,1)$

Question

I am interested in computing the value of a complex expression (see here) containing the following building block

$$ \prod_{j=1}^{n}\frac{j}{\lceil{\alpha j}\rceil}, $$

and its partial product starting from $j=i$, where $\alpha\in(0,1)$ is a parameter. I am interested in the behavior as $n\rightarrow\infty$. While for all $\alpha\in(0,1)$, the above expression is unbounded as $n$ grows, I would want to understand the asymptotic behavior of this infinite product parametrized by $\alpha$.

• Has anyone ever encountered similar expressions?
• Any ideas of how to get the asymptotic?
• Also pointers to connections with other functions in combinatorics/number theory would be useful.

Note: the fundamental difficulty arises from understanding the behavior of the term on the denominator: $$ \lceil\alpha\rceil \lceil2\alpha\rceil \lceil3\alpha\rceil ...\lceil n\alpha\rceil, $$ as the numerator can be simply written as $(n+1)!$

Thank you :-)

Answer 1

Hint:

$$\alpha j\le\lceil \alpha j\rceil<\alpha j+1$$

and

$$\alpha^nn!\le\prod_{j=1}^n\lceil\alpha j\rceil<\prod_{j=1}^n(\alpha j+1).$$

Then $$\prod_{j=1}^n\left(\alpha+\dfrac1j\right)<\prod_{j=1}^n\left(1+\dfrac1j\right)=n+1$$

so

$$\alpha^nn!\le\prod_{j=1}^n\lceil\alpha j\rceil<(n+1)!$$

Answer 2

I looked at a simple case, and it's already fairly subtle. For example, say $\alpha=0.5$. It is not difficult to show that $\prod_{j=1}^n \lceil \alpha j \rceil = (n/2)! (n/2)!$ when $n$ is even. In this case, using Stirling approximants, we get $$ \prod_{j=1}^n \frac{j}{\lceil \alpha j \rceil}=\frac{n!}{(n/2)!(n/2)!}\sim \frac{2^{n+1}}{\sqrt{2\pi n}}, $$ also verified numerically. The reasoning (based on the explicit computation of the product of the ceilings) can be extended with a bit of care to any $\alpha=1/q$ with $q$ integer, but brakes down already with $\alpha = p/q$, and $p,q$ integers.