Theorem: Let $\sum_{n=0}^{\infty}a_nz^n$ be a power series with radius of convergence $R$ and $f:U\to \mathbb{C}$, \begin{equation}f(z)=\sum_{n=0}^{\infty}a_nz^n\end{equation} where $U=B(0,r)$, $0
. Then $f$ is holomorphic at $0$ and \begin{equation}f^{\prime}(z)=\sum_{n=1}^{\infty}na_nz^{n-1} \end{equation}
Question: I have been reading the proof of this from here http://www.math.iitb.ac.in/~ars/revbook.pdf pg 80 and I don't understand how the last two lines in that page are derived.
Proof:
After we show that both series have the same radius of convergence $R$ let $g:U\to \mathbb{C}$, \begin{equation}g(z)=\sum_{n=1}^{\infty}na_nz^{n-1} \end{equation} Choose $h$ with $0<\left|h\right|
I don't understand how the bold face sentence implies the last inequality. I know that $\sum_{n=N}^{\infty}n\left|a_n\right|r^{n-1}$ converges but how is it less than an arbitary $\epsilon$?.