For $\alpha, \beta, \gamma \in \mathbb{C}$ consider the bi-infinite tridiagonal Laurent operator $T$ with $\beta $ on the diagonal given by.
\begin{pmatrix} \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \\ \dots & \alpha & \beta & \gamma & 0 & 0 & 0 & \dots \\ \dots & 0 & \alpha & \beta & \gamma & 0 & 0 & \dots\\ \dots & 0 & 0 & \alpha & \beta & \gamma & 0 & \dots\\ \dots & 0 & 0 & 0 & \alpha & \beta & \gamma & \dots \\ \dots & \dots & \dots & \dots & \dots & \dots & \dots & \dots \\ \end{pmatrix}
General theory tells us that $T$ is invertible if and only the symbol curve (which is in this case is an ellipsis) given by $ \{ z \in \mathbb{T} \mid \frac{\alpha}{z} + \beta + z \gamma \} $ does not enclose $0$.
Suppose that this is the case and let $e_0$ denote the a unit vector in the standard basis. Then what is the value $ \langle e_0 , T^{-1} e_0 \rangle ? $
As we explain in the Appendix here: https://arxiv.org/abs/2206.09879 this is a standard calculation: "First, $\sigma(T)$ is the image of the symbol curve $ a(z) = \alpha z^{-1} + \beta + \gamma z $ for $z \in \mathbb{T}$. Since $T$ is invertible it holds that $ a(z) \neq 0$ for all $z \in \mathbb{T}$ and therefore, the symbol curve of the inverse is given by \begin{align*} \frac{1}{a(z)} = \frac{1}{ \frac{\alpha}{ z} + \beta + \gamma z } = \frac{z}{ \alpha + \beta z + \gamma z^2} = \frac{z}{ \gamma( \frac{\alpha}{\gamma} + \frac{\beta}{\gamma} z + z^2) }. \end{align*} We can rewrite the denominator $ \gamma ( z - \lambda_+) (z- \lambda_-) $ with \begin{align*} \lambda_{\pm} = \frac{ - \beta}{2 \gamma} \pm \sqrt{ \left( \frac{\beta}{2 \gamma} \right)^2 - \frac{\alpha }{\gamma} }. \end{align*} Notice that \begin{align*} \lambda_+ \lambda_- = \frac{\alpha}{\gamma} \text{ , } \lambda_+ + \lambda_- = - \frac{\beta}{\gamma}, \text{ and } \lambda_+ - \lambda_- = 2 \sqrt{ \left( \frac{\beta}{2 \gamma} \right)^2 - \frac{\alpha }{\gamma} }. \end{align*} Now, assuming that $\vert \lambda_2 \vert < 1< \vert \lambda_1 \vert $ (where $\{\lambda_2, \lambda_1 \} = \{\lambda_+, \lambda_- \}$ has implications on how we write this up as a geometric series): \begin{align*} \frac{1}{a(z)} & = \frac{z}{ \gamma ( z - \lambda_+) (z- \lambda_-) } = \frac{z}{ \gamma ( \lambda_1 - \lambda_2)} \left( \frac{1}{z-\lambda_1} - \frac{1}{z-\lambda_2} \right) \\ & = \frac{z}{ \gamma ( \lambda_1 - \lambda_2)} \left( - \frac{1}{\lambda_1} \frac{1}{1- \frac{z}{\lambda} } - \frac{1}{z} \frac{1}{1-\frac{\lambda_2}{z} } \right) \\ & = \frac{z}{ \gamma ( \lambda_1 - \lambda_2)} \left( - \frac{1}{\lambda_1} \sum_{n=0}^\infty (\frac{z}{\lambda_1})^n - \frac{1}{z} \sum_{n=0}^\infty (\frac{\lambda_2}{z})^n \right) \\ & = \frac{1}{ \gamma ( \lambda_2 - \lambda_1)} \left( \sum_{n=1}^\infty (\frac{z}{\lambda_1})^n + \sum_{n=0}^\infty (\frac{\lambda_2}{z})^n \right). \end{align*}"
Then one can read of coefficients to obtain the answer.