Does this extreme huge size determinant converge?

Question

Introduction:

One day I calculated the value of determinant which is like Hilbert matrix $H_{n}^{p} \in \bf{R}^{\it{n \times n}}$using my computer. The determinant is defined below.

$$ \det{(H_{n}^{p})}= \left| \begin{array}{ccccc} 1 & {2^{p}} & {3^{p}} & \dots & {n^{p}} \\ &&&&\\ {2^{p}} & {3^{p}} & {4^{p}} & \dots & {{(n+1)}^{p}} \\ &&&&\\ {3^{p}} & {4^{p}} & {5^{p}} & \dots & {{(n+2)}^{p}} \\ &&&&\\ \vdots & \vdots & \vdots & \ddots & \vdots \\ &&&&\\ {n^{p}} & {{(n+1)}^{p}} & {{(n+2)}^{p}} & \dots & {{(2n-1)}^{p}} \end{array} \right| $$

This figure is the result of calculation. The horizontal axis is size of matrix as $n$ ,and the vertical axis is value of the determinant. The calculation range is between $p=[-2,18]$ , $n=[1,200]$, and the calculative separation is $\Delta p=0.1$.

But moreover, I want to know whether the convergence range of $p$ exists or not exactly. However I think this mere question is very difficult. So could you consider this problem when you have time?

Problem:

If $\det{(H_{\infty}^{p})}$ converges to a constant value, estimate the range of $p$.

Answer 1

Here is a partial answer, edited to fit the newly posed problem.

The determinant of a matrix corresponds to the product of its eigenvalues. If $p=0$ the matrix is singular and so one of its eigenvalues is zero, hence the determinant exists and is zero.

If $p<-1$ then the sum of the elements on the main diagonal converges to a finite value as $n\to\infty$. This sum is called the trace. The trace equals also the sum of all eigenvalues. Hence, as $n$ increases all eigenvalues must eventually be smaller than $1$ in absolute values and therefore, their product converges to zero and so does the determinant.

Finally, I guess that we have also convergence for $p\in [-1,0]$ but that for $p>0$ the determinant diverges to infinity (maybe the Leibniz formula for a determinant can prove useful).

Answer 2

Some results.

If $p$ is a positive integer, then $\det(H_n^p)=0$ when $n\geq p+2$.

If $p=-1$, then we obtain the Hilbert matrix and $\log(\det(H_n^{-1}))\sim -n^2\log(4)$.

In the sequel, I work with $300$ significative digits. More generally, it seems that, for every fixed $p\in\mathbb{R}$, $\log(|\det(H_n^p)|)=O(n^2)$ and $\det(H_n^p)$ tends to $0$ when $n$ tends to $\infty$. Note that $\det(H_n^P)$ may be negative.

Although the OP does not seem to be interested by my post, I give the following example:

With $700$ digits, let $p=21.5$: $|\det(H_n^p)|$ increases until $n=29$ with $|\det(H_{29}^p)|\approx 10^{492}$. After, it decreases $|\det(H_{55}^p)|\approx 10^{27},|\det(H_{56}^p)|\approx 10^{-9}, |\det(H_{70}^p)|\approx 10^{-634}$.

Now, it's no more my business...