convergence in ${\mathfrak{S}_{1}}$ is properly justified — can be used to evaluate eigenvalues of the operator $L = L^0 + W$.

Under proper conditions, if $n \geq N_*$, the operator $L$ has the only eigenvalue $\lambda_n$ in $H_n$; moreover, $\lambda_n$ is simple and $\lambda_n \in \mathcal{D}_n$. Therefore, both of the projections $$\begin{aligned}
P_n^0 & = \frac{1}{2\pi i} \int\limits_{{\partial (\mathcal{D}_n )}} R^0(z) \, dz = {\left\langle \cdot, h_n \right\rangle} h_n \label{eq:pnbase}\\
\intertext{and}
P_n   & = \frac{1}{2\pi i} \int\limits_{{\partial (\mathcal{D}_n )}} R(z) \, dz = {\left\langle \cdot, \psi_n \right\rangle} \phi_n \label{eq:pn}\end{aligned}$$ are of rank $1$. \[In , $\phi_n$ is an eigenfunction of $L$ and $\psi_n$ is an eigenfunction of $L^* = L^0 + W^*$, with an eigenvalue $\mu_n = \overline{\lambda}_n$ in $\mathcal{D}_n$. We will not use this specific information so nothing more is explained now.\]

Therefore, $$\begin{aligned}
\operatorname{Trace}P_n^0 = \operatorname{Trace}P_n &= 1, \label{eq:trabase}\\
\operatorname{Trace}\frac{1}{2\pi i} \int\limits_{{\partial (\mathcal{D}_n )}} (R(z) - R^0(z)) \, dz &= 0. \label{eq:trafull}\end{aligned}$$ and $$\begin{aligned}
z_n & = \operatorname{Trace}\frac{1}{2\pi i} \int\limits_{{\partial (\mathcal{D}_n )}} z R^0(z) \, dz = 2n + 1, \label{eq:lambase}\\
\lambda_n & = \operatorname{Trace}\frac{1}{2\pi i} \int\limits_{{\partial (\mathcal{D}_n )}} z R(z) \, dz, \label{eq:lamfull}\end{aligned}$$ So imply $$\begin{aligned}
\lambda_n - z_n &= \operatorname{Trace}\frac{1}{2\pi i} \int\limits_{{\partial (\mathcal{D}_n )}} (z - 