On the third edition of Ahlfors' Complex Analysis, page 40 Theorem 2 it states: we conclude that
\begin{equation*}
\left|\frac{R_n(z)-R_n(z_0)}{z-z_0}\right| \leq \sum_{k=n}^{\infty} k|a_k|\rho^{k-1}
\end{equation*}
The expression on the right is the remainder term in a convergent series.
I have trouble in understanding that $\sum_{k=n}^{\infty} k|a_k|\rho^{k-1}$ is convergent.
We know that $1/{\rho}>1/R=\lim_{n\rightarrow\infty} \sup \sqrt[n]{|a_n|}$. So, there exists an $N$ such that $1/\rho>\sqrt[k]{|a_k|}, |a_k|<\frac{1}{\rho ^k}$ for all $k\geq N$. If $n\geq N$, we have
\begin{equation*}
\sum_{k=n}^{\infty} k|a_k|\rho^{k-1}\leq\sum_{k=n}^{\infty} k\frac{1}{\rho^k}\rho^{k-1}=\sum_{k=n}^{\infty}\frac{k}{\rho}
\end{equation*}
Since $\sum_{k=n}^{\infty}\frac{k}{\rho}$ is not convergent, we cannot conclude that $\sum_{k=n}^{\infty} k|a_k|\rho^{k-1}$ is convergent. How to prove $\sum_{k=n}^{\infty} k|a_k|\rho^{k-1}$ is convergent?
Thanks for eulersgroupie's answer. I'd like to add more details. The radius of convergence of $\sum_{1}^{\infty}na_n z^{n-1}$ is $R$. Since $\rho <R$, $\sum_{1}^{\infty}na_n \rho^{n-1}$ is absolutely convergent. So, $\sum_{1}^{\infty}n|a_n| \rho^{n-1}$ is convergent. Finally, $\sum_{k=n}^{\infty} k|a_k|\rho^{k-1}$ is convergent.
Martin R's answer likes a magic and I really appreciate it. It solves my question directly.
This a question from Ahlfors' Complex Analysis. I show you the whole context of the question below. I have trouble understanding the statement with red line.
Best Answer
$1/\rho>\sqrt[k]{|a_k|}$ for sufficently large $k$ is correct, but not good enough to prove the convergence.
Choose $r$ such that $\rho < r < R$, then $1/{\rho}>1/r > 1/R=\lim_{n\rightarrow\infty} \sup \sqrt[n]{|a_n|}$ so that $1/r>\sqrt[k]{|a_k|}$ for sufficiently large $k > N$. Then $$ \sum_{k=n}^{\infty} k|a_k|\rho^{k-1}\leq \sum_{k=n}^{\infty} \frac{k}{\rho} \left( \frac{\rho}{r}\right)^k $$ and that series is convergent.