in norm as $t\to\pm\infty$ but there exists a polynomial $\mathfrak{p}_k$ with positive coefficients such that $$\deg \mathfrak{p}_k =k$$ and for every $t\in\mathbb{R}$ we have $$\left\|e^{A_k t}\right\| \leq \mathfrak{p}_k (|t|).$$ So $\left\{e^{A_k t} \right\}_{t\in \mathbb{R}}$ belongs to the class of polynomially bounded $C_0$-groups. Hence the growth bound $\omega_{0,k}$ of the $C_0$-group $\left\{e^{A_k t} \right\}_{t\in \mathbb{R}}$ equals to zero, i.e. $$\omega_{0,k}=\lim\limits_{t\to\pm\infty} \frac{\ln\left\|e^{A_k t} \right\|}{t}=0,$$ see also Corollary 14 in [@Sklyar3]. Therefore the first part of the Theorem follows from the well-known representation of the resolvent of the generator in the form of the Laplace transform of the $C_0$-semigroup (group): $$\label{representation res}
\left(A_k- \lambda I \right)^{-1}=\int\limits_{0}^{\infty} e^{-\lambda t} e^{A_k t} dt,\quad \left|\Re(\lambda)\right|>0,$$ see, e.g., Theorem 11 in [@Dunford], see also Theorem 2.1 of [@Eisner1].

2\. Follows from Theorem 2.6 of [@Eisner2].

3\. Follows from Theorem 3 of [@Malejki].

Consider the case $k=1$ in Theorem \[res asympt\]. Then the conclusion of the Theorem \[res asympt\] follows from the proof of the Theorem \[Expl\]. Indeed, by (\[alambda\]) we have that for all $\lambda \in  \rho\left(A_1\right)=\mathbb{C}\setminus \{i f(n)\}_{1}^{\infty}$ $$\left\|\left(A_1- \lambda I \right)^{-1}\right\| \leq \sqrt{\frac{2M}{m}}\frac{\sqrt{\left(\inf\limits_{n\in \mathbb{N}} |if(n)-\lambda|\right)^2+4 \left(\sup\limits_{n\in\mathbb{N}} n|\Delta f(n)| \right)^2}}{\left(\inf\limits_{n\in \mathbb{N}} |if(n)-\lambda|\right)^2},$$ which obviously leads to estimates (\[r est\]) and (\[r est 2\]). Thus, in general, 