+ 1), \quad k = 0, 1, \dotsc,
\label{eq:ell0}$$ and $W = (w_{jk})_0^{\infty}$ a matrix such that for some $\alpha > 0$ and $C_0 > 0$, $$\vert w_{jk} \vert \leq \frac{C_0}{(1 + j)^{\alpha} (1 + k)^{\alpha}} \label{eq:wcond}$$ Then the quadratic-form method [see @ReedSimon Section VII.6] leads to the definition of the closed operator $$L = L^0 + W \label{eq:elldef}$$ with a dense domain — see details in [@MiSi]. Let us recall some facts, introduce notations and explain a few elementary but important inequalities.

To adjust our constructions to the set of eigenvalues of the unperturbed operator , let us define strips $$\begin{aligned}
H_n & = {\left\lbrace z \in {\mathbb{C}}: \vert {\operatorname{Re}}z - z_n \vert \leq 1 \right\rbrace}, \quad n \geq 1\\
H_0 & = {\left\lbrace z \in {\mathbb{C}}: {\operatorname{Re}}z - z_0 \leq 1 \right\rbrace}
\end{aligned} \label{eq:hns}$$ and the squares $$\mathcal{D}_n = {\left\lbrace z \in H_n: \vert {\operatorname{Re}}z - z_n \vert \leq \frac{1}{2}, \vert {\operatorname{Im}}z \vert \leq \frac{1}{2} \right\rbrace}, \quad n \geq 0  \label{eq:dn}$$ around eigenvalues $ {\left\lbrace z_n \right\rbrace}_{n = 1}^{\infty} = {\left\lbrace 2n + 1 \right\rbrace}_{n = 0}^{\infty}$ in $H_n$.

The resolvent $$R(z) = (z - L^0 - W)^{-1} \label{eq:res}$$ of the operator is well-defined in the right half-plane 