Let $A$ be a nonnegative matrix.
The resolvent $R(\lambda) = (\lambda I − A)^{−1}$ is analytic except at its spectrum (the eigenvalues). It must be analytic on the annulus $|λ| > ρ$ and must have at least one singularity somewhere on the spectral radius circle $|λ| = ρ$.
The Neumann series for the resolvent $$R(\lambda) = \sum _{k=1}^{\infty} \frac{A^{k-1}}{\rho^k}$$ will converge down to the first singularity of $R(λ)$.
I could not understand the meaning of the last statement. Please explain.
There is a unitary $U$ such that $A=U^*DU$ where $D$ is a diagonal matrix with eigenvalues $\rho = \lambda_1 \ge \lambda_2 \ge \cdots \ge \lambda_k \ge 0$. Then $$ (\lambda I-A)^{-1}=U^*\begin{pmatrix}\frac{1}{\lambda-\lambda_1} & 0 & 0 & \cdots & 0 \\ 0 & \frac{1}{\lambda-\lambda_2} & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \frac{1}{\lambda-\lambda_n}\end{pmatrix}U. $$ Then, for $|\lambda| > \rho$, you can expand \begin{align} \frac{1}{\lambda-\lambda_k}& = \frac{1}{\lambda}\frac{1}{1-\frac{\lambda_k}{\lambda}}=\sum_{m=0}^{\infty}\frac{\lambda_k^m}{\lambda^{m+1}}. \end{align} So, $$ (\lambda I-A)^{-1}=\sum_{m=0}^{\infty}\frac{1}{\lambda^{m+1}} U^*\begin{pmatrix}\lambda_1^{m} & 0 & 0 & \cdots & 0 \\ 0 & \lambda_2^m & 0 & \cdots & 0 \\ \vdots & \vdots & \vdots & \ddots & \vdots \\ 0 & 0 & 0 & \cdots & \lambda_n^m \end{pmatrix}U, \;\;\; |\lambda| > \rho. $$ This can be written as $$ (\lambda I - A)^{-1}=\sum_{m=0}^{\infty}\frac{1}{\lambda^{m+1}}A^{m}, \;\;\; |\lambda| > \rho. $$