I have the following matrix
$$H=\begin{bmatrix} 1 & \frac{1}{2} & \cdots & \mbox{ad}\ +\infty\\ \frac{1}{2} & \frac{1}{3} & \cdots & \mbox{ad}\ +\infty\\ \vdots & \vdots & \ddots\\ \ & \ & \mbox{ad}\ +\infty \end{bmatrix}$$
which is a Hilbert matrix of order $\infty$. My problem is to find the largest eigenvalue $\lambda_{max}$ of $H$ and find an eigenvector corresponding to $\lambda_{max}$. I do not think the conventional way of finding eigenvalues and eigenvectors is going to help me here. Otherwise, I am not sure how to proceed to solve this question. Please help.
$H$ does not have $\pi$ as an eigenvalue (interpreting $H$ as operating on $\ell^2$). See for instance Theorem 323 of Hardy, Littlewood and Polya, "Inequalities" where it is shown that for a non-zero $x$ in $\ell^2$, $x^THx<\pi\|x\|^2$ strictly. But there are eigenvalues arbitrarily close to $\pi$ since if $x_r=1/\sqrt r$ for $r=1,\ldots,N$ and $x_r=0$ for $r>N$ then $x^THx/\|x\|^2$ tends to $\pi$ as $N\to\infty$.