$A_{nm}$ ($n,m = 0, 1, 2, \ldots$) is a symmetric, tridiagonal matrix. The diagonal elements are $A_{nn} = a_n = n + 1$, and the off-diagonal elements are $A_{n,n+1} = A_{n+1,n} = b_n = \lambda \sqrt{\lfloor\frac{n+2}{2}\rfloor}$, where $\lambda$ is a real parameter. $$A = \begin{bmatrix}1 &\lambda& 0 &0 &0 &0 &0&\cdots \\\ \lambda & 2 & \lambda & 0 &0 &0&0 & \cdots \\\ 0 & \lambda & 3 & \lambda \sqrt{2} & 0 &0 &0&\cdots\\\ 0 &0 &\lambda \sqrt{2} & 4 & \lambda \sqrt{2} & 0 &0& \cdots\\\ 0 & 0 & 0 & \lambda \sqrt{2} & 5 & \lambda \sqrt{3} &0&\cdots \\\ 0 &0 &0 &0 & \lambda \sqrt{3} & 6 &\lambda \sqrt{3}& \cdots \\\ 0 &0 &0 &0 &0 & \lambda \sqrt{3} &7& \cdots \\\ \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \vdots & \ddots\end{bmatrix}$$
For any $\lambda$, the spectrum of this matrix is bounded from below. I would like to know its lowest eigenvalue $\epsilon_0$ as a function of $\lambda$. For $\lambda = 0$, the matrix is diagonal and the eigenvalue of interest is $1$. For small $\lambda$, I have worked through perturbation series sums and found $\epsilon_0 (\lambda) = \sum_{n=0} c_n \lambda^n = 1 - \lambda^2 + \lambda^4/2 - \lambda^6 2/3 + \lambda^8 79/72 - \lambda^{10} 274 / 135 + \cdots$, but I cannot identify any pattern to this. The Taylor series has a finite radius of convergence, which by an analysis of the partial series I have shown numerically to be near $\vert \lambda \vert = 0.6134\ldots$.
Is there a closed-form expression for $\epsilon_0(\lambda)$? If not, is there a simple expression for the Taylor series coefficients $c_n$ and its radius of convergence?