$\renewcommand{\Re}{\operatorname{Re}}$ $\renewcommand{\Im}{\operatorname{Im}}$ $\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\bC}{\mathbb{C}}$ $\newcommand{\bZ}{\mathbb{Z}}$ $\newcommand{\const}{\operatorname{const}}$
See Definition 3.1.1.
We consider complex-analytic solutions to the equation \begin{equation} u^{(n)}(z)+ p_{n-1} (z)u^{(n-1)}(z)+\ldots+p_1 (z)u'(z)+p_0(z)u(z)=0 \label{eq-3.2.1} \end{equation} where $p_k(z)$ are complex-analytic functions at $z_0\in \bC$.
Assume for a sake of simplicity notations that $z_0=0$ (we can always reach it by the change of variables $\zeta=z-z_0$). Then \begin{equation} p_k(z)= \sum_{l\ge 0} p_{kl}z^l \label{eq-3.2.2} \end{equation} and we are looking for solution \begin{equation} u(z)=\sum_{l\ge 0} u_l z^l. \label{eq-3.2.3} \end{equation} Plugging into (\ref{eq-3.2.1}) we get \begin{equation*} \sum_{l\ge n} u_l \frac{l!}{(l-n)!} z^{l-n} + \sum_{0\le k\le n-1}\ \sum_{m\ge 0}\ \sum_{l\ge k} p_{km}u_{l-k} \frac{l!}{(l-k)!} z^{l-k+m}=0 \end{equation*} or changing index in summation ($l:=l+n$ in the first term, $l:=l+k-m$ in the other terms) we get \begin{equation*} \sum_{l\ge 0} u_{l+n} \frac{(l+n)!}{l!} z^{l} + \sum_{0\le k\le n-1}\ \sum_{m\ge 0}\ \sum_{l\ge m} p_{km}u_{l-k-m} \frac{(l+k-m)!}{(l-m)!} z^{l}=0 \end{equation*} or equivalently \begin{equation*} u_{l+n} \frac{(l+n)!}{l!} + \sum_{0\le k\le n-1}\ \sum_{m\le l} p_{km}u_{l+k-m} \frac{(l+k-m)!}{(l-m)!} =0 \end{equation*} and finally \begin{equation} u_{l+n} = -\frac{l!}{(l+n)!} \sum_{0\le k\le n-1}\ \sum_{m\le l} p_{km}u_{l +k-m} \frac{(l+k-m)!}{(l-m)!} \label{eq-3.2.4} \end{equation} which defines $u_l$ for $l=n,n+1,\ldots$ recurrently as long as $u_0,\ldots, u_{n-l}$ could be chosen arbitrarily.
Theorem 1. If series (\ref{eq-3.2.2}) converge as $|z|< R$ and $u_l$ for $l=n,n+1,\ldots$ are defined by (\ref{eq-3.2.4}) then series (\ref{eq-3.2.3}) converges as $|z|< R$.
Corollary 1. If series (\ref{eq-3.2.2}) converge for all $z$ and $u_l$ for $l=n,n+1,\ldots$ are defined by (\ref{eq-3.2.4}) then series (\ref{eq-3.2.3}) converges for all $z$.
Remark 1. Such functions are called entire.
We sketch the proof. We can rewrite equation as the first order system \begin{equation} U'(z)=\Lambda (z) U(z) \label{eq-3.2.5} \end{equation} where $U=(u\ u'\ \ldots \ u^{(n-1)})^T$, $^T$ means a transposed matrix and $\Lambda(z)$ is analytic as $|z|< R$.
Definition 1. Let us consider two series: $\sum a_k z^k$ and $\sum_k b_kz^k$. We say that $\sum_k b_kz^k$ dominates $\sum_k a_kz^k$, $\sum_k b_kz^k\ggg \sum_k b_kz^k$ if $|a_k|\le b_k$ for all $k$. Attention! Norm is only in the left! For vector- or matrix- valued functions it should be for each component.
One can prove easily that if $U$ is solution of (\ref{eq-3.2.5}) and $V$ is solution of the similar system \begin{equation} V'(z)=\Lambda_1 (z) V(z) \label{eq-3.2.6} \end{equation} with matrix $\Lambda_1(z)$ which dominates $\Lambda(z)$ and if $|U_j (0)|\le V_j(0)$ then $U(z)\lll V(z)$.
But if series (\ref{eq-3.2.2}) converges as $|z|< R$ then $\Lambda(z)\lll M E(r-z)^{-1}$ for any $r< R$ and $M=M_r$. Here $E$ is a matrix with all elements $1$. If we add all equations in system (\ref{eq-3.2.6}) we get a single equation \begin{equation} v'=Mn(r-z)^{-1} v \label{eq-3.2.7} \end{equation} with $v=V_1+\ldots +V_n$. Solution of this equation is $v(z)=v(0) + Mn (\ln r- \ln (r-z))$ which is analytic as $|z|< r$ and therefore it Taylor decomposition at $z=0$ converges in the disk $\{z:\,|z|< r\}$. Since $v(z)\ggg u(z)$ Taylor decomposition of $u(z)$ at $z=0$ converges in the same disk.