Determinant with factorials

Question

Let $D_n$ be the determinant $$\begin{vmatrix} \frac1{(n+1)!}&\cdots&\cdots&\frac1{1!}\\ \vdots&&&\vdots\\ \vdots&&&\vdots\\ \frac1{(2n+1)!}&\cdots&\cdots&\frac1{(n+1)!} \end{vmatrix}$$ Can one determine the value or at least an equivalent of $D_n$ when $n\to+\infty$? Thanks in advance for any answer.

Answer 1

There is a simple explicit expression for this determinant. But the proof is not that simple. Here are the ideas:

I rewrite it has a Hankel determinant of order $n$, it just change up to a sign $(-1)^{n(n-1)/2}$ and a shift.

First consider the determinant:

$$D_n(x)=\begin{vmatrix} \frac1{1!}x&\cdots&\cdots&\frac1{n!}x^n\\ \vdots&&&\vdots\\ \vdots&&&\vdots\\ \frac1{n!}x^n&\cdots&\cdots&\frac1{(2n-1)!}x^{2n-1} \end{vmatrix}. $$

Then derive it $n^2$ times, it yields with some multiplicity, the triangular determinant: $$\begin{vmatrix} 0 & \dots & & 0 & 1\\ \vdots & & & 1 & x\\ \vdots & & & & \vdots\\ 0 & 1 & \dots & & \frac1{(n-2)!}x^{n-2}\\ 1 & x & \dots & \frac1{(n-2)!}x^{n-2} & \frac1{(n-1)!}x^{n-1}\\ \end{vmatrix}=(-1)^{n(n-1)/2}. $$

The multiplicity is the multidimensionnal ballot number from $(0,1,\dots,n-1)$ to $(n,n+1,\dots,2n-1)$: $$\frac{(n^2)!(n!!)^2}{(2n)!!}$$ where $n!!=\prod_{k=0}^{n-1}(k!)$. (Ref. in Nathanson, Additive Combinatorics)

So we deduce that $$D_n(x)=(-1)^{n(n-1)/2}\frac{(n^2)!(n!!)^2}{(2n)!!}\frac1{(n^2)!}x^{n^2}=(-1)^{n(n-1)/2}\frac{(n!!)^2}{(2n)!!}x^{n^2},$$

and finally $$D_n=(-1)^{n(n-1)/2}\frac{(n!!)^2}{(2n)!!}=(-1)^{n(n-1)/2}\prod_{i=0}^{n-1}\frac{i!}{(n+i)!},$$ and therefore: $$|D_n|\leq \frac1{(n!)^n}$$

I skipped some details and there may be simpler proofs, but you got the expression that I checked for $n=2,3$.