It suffices to show that the power series
\begin{equation*}
g(z)=a_m+a_{m-1}(z-z_0)+\cdots =\sum_{n=0}^\infty a_{m+n}(z-z_0)^n
\end{equation*}
converges for all \(z\in B_R(z_0)\text{,}\) since then \(g\) is analytic on \(B_R(z_0)\) and we have
\begin{equation*}
(z-z_0)^mg(z)=\sum{n=0}^\infty a_{m+n}(z-z_0)^{m+n}=\sum_{n=m}^\infty a_n(z-z_0)^n=f(z)\text{.}
\end{equation*}
Clearly \(g(z)\) converges for \(z=z_0\text{.}\) Assume now that \(z\in B_R(z_0)-\{z_0\}\text{.}\) Let \(g_k\) and \(f_k\) be the \(k\)-th partial sums of \(f(z)\) and \(g(z)\text{,}\) respectively. We will show that \(g_k(z)\to f(z)/(z-z_0)^m\) and thus that \(g(z)=f(z)/(z-z_0)^m\) converges. First observe that
\begin{align*}
\abs{g_k(z)-\frac{f(z)}{(z-z_0)^m}} \amp = \abs{\frac{g_k(z)(z-z_0)^m-f(z)}{(z-z_0)^m}}\\
\amp = \abs{\frac{g_k(z)(z-z_0)^m-f_k(z)+(f_k(z)-f(z))}{(z-z_0)^m}} \\
\amp = \abs{\frac{f_k(z)-f(z)}{(z-z_0)^m}} \amp (g_k(z)(z-z_0)^m=f_k(z))\text{.}
\end{align*}
Let \(r=\abs{z-z_0}\text{.}\) Given any \(\epsilon > 0\text{,}\) since \(f_k\to f\text{,}\) we can find an integer \(N\geq 0\) such that \(\abs{f_k(z)-f(z)}< \epsilon r^n\text{.}\) It follows that
\begin{equation*}
\abs{g_k(z)-\frac{f(z)}{(z-z_0)^m}}=\abs{\frac{f_k(z)-f(z)}{(z-z_0)^m}}< \frac{\epsilon r^n}{r^n}=\epsilon\text{.}
\end{equation*}
This proves \(g(z)=\lim_{k\to \infty}g_k(z)=f(z)/(z-z_0)^m\text{,}\) as claimed.
