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Let $x_k>0$ be a increasing sequence of real numbers, such that $$\sum_0^\infty\frac1{x_k}<+\infty.$$ Let us form the (infinite) Hilbert matrix $A\in{\bf Sym}({\mathbb N};{\mathbb R})$ with $$a_{ij}:=\frac1{x_i+x_j}\,.$$ Mind that $A$ is trace class, thus the eigenvalues $\lambda_0>\lambda_1\ge\cdots>0$ are summable.

What is known about the asymptotics of the eigenvalues $\lambda_j$, in terms of that of $x_k$ ? For instance, what do we know about the spectrum of $A$ when $x_k=(k+1)^2$ ?

Another form of the question is the behaviour, as $N\to+\infty$, of the spectrum of the truncated matrix $A_N$ in which $0\le i,j,\le N$.

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    $\begingroup$ Are you willing to change $\frac{1}{k^2+n^2}$ to $\frac{1}{(k+n)^2}$? Because then such matrices are known as Hankel matrices, and I found in a paper by Pushnitski and Yafaev a reference to a paper by Widom, in which he apparantly proved asymptotics $\lambda_n = \exp(-2\pi \sqrt{n} + o(\sqrt{n}))$ and for general $\frac{1}{(k+n)^\gamma}$, $\gamma > 1$, $\lambda_n = \exp(-\pi \sqrt{2\gamma}\sqrt{n} + o(\sqrt{n}))$. $\endgroup$
    – Aleksei Kulikov
    Commented Nov 30, 2022 at 11:14
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    $\begingroup$ I do not see why $A$ is trace class in this generality. Is this always true and I am missing something? $\endgroup$
    – Giorgio Metafune
    Commented Nov 30, 2022 at 17:54
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    $\begingroup$ @AlekseiKulikov I'm a bit suspicious about your formula for general $\gamma$: when $\gamma\to 1$, it still provides some fast rate of decay, while the operator gets certainly worse and worse and in the limit there is no decay at all. Of course, it may happen that it isjust $o()$ that deteriorates, but that would be a bit surprising. $\endgroup$
    – fedja
    Commented Dec 3, 2022 at 4:16
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    $\begingroup$ @fedja yes, you're absolutely right. I copied the (incorrect) formula from Pushnitski and Yafaev, but in the Widom's paper he uses $\gamma-1$ instead of $\gamma$. So for $\frac{1}{(k+n)^2}$ it should be $\lambda_n = \exp(-\pi \sqrt{2}\sqrt{n}+o(\sqrt{n}))$. $\endgroup$
    – Aleksei Kulikov
    Commented Dec 3, 2022 at 11:58
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    $\begingroup$ @AlekseiKulikov Based on a quick back of envelope computation, I would rather expect $(\gamma-1)/\gamma$ but Hankel may be a bit different from Hilbert, so I'll not insist here :-) $\endgroup$
    – fedja
    Commented Dec 3, 2022 at 12:08

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