I am wondering if there is a way to get an explicit expression for the eigenvalues (and possibly the eigenvectors) of a symmetric tridiagonal matrix with the following peculiar structure: $$ \left( \begin{array}{ccccc} 0 & k\sqrt{1} & 0 & 0 & \dots \\ k\sqrt{1} & 1 & k\sqrt{2} & 0 & \dots \\ 0 & k\sqrt{2} & 2 & k\sqrt{3} & \dots \\ 0 & 0 & k\sqrt{3} & 3 & \dots \\ \vdots & \vdots & \vdots & \vdots & \ddots \end{array} \right) $$ in the limit where the size of the matrix goes to infinity. The context of this problem is quantum physics, and by some arguments that are shown here, the answer should be something like $\lambda_n = n - k^2$.
I know that the characteristic polinomial can be written recursively as $\lim_{n\to\infty}p_n(\lambda)$, where $$ p_n(\lambda)=a_n(\lambda) p_{n-1}(\lambda) - b_{n-1}^2 p_{n-2}(\lambda), \;\;\;\;\; p_{0}(\lambda) = 1 \;\;\;\;\; p_{-1}(\lambda) = 0, $$ $a_n(\lambda)$ being the diagonal term, so $a_n=n-1-\lambda$ and $b_n$ being the off diagonal term, thus $b_n=k\sqrt{n}$, but I am stuck here.
Can anyone help me with this? Thanks!