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 $U=B(0,r)$, $0<r<R$. 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 $0<\left|h\right|<r-\left|z\right|$. Then $z+h\in U$. If $u_n(h)=a_n\frac{(z+h)^n-z^n}{h}-na_nz^{n-1}$ then, \begin{equation}u_n(h)=a_n\frac{(z+h)^n-z^n}{h}-na_nz^{n-1}=a_n\sum_{k=0}^{n-1}(z+h)^{n-k-1}z^k-nz^{n-1} \end{equation} Since $\left|z\right|<r$ and $\left|z+h\right|<r$, \begin{equation}\left|u_n(h)\right|\le\left|a_n\right|\sum_{k=0}^{n-1}r^{n-1}+nr^{n-1}=2n\left|a_n\right|r^{n-1} \end{equation} Since the radius of convergence of series $\sum_{n=1}^{\infty}na_nz^{n-1}$ is $R>r$ we have that $\exists N\in \mathbb{N}$ so that \begin{equation}\left|\sum_{n=N}^{\infty}u_n(h)\right|\le \sum_{n=N}^{\infty}\left|u_n(h)\right| \le 2\sum_{n=N}^{\infty}n\left|a_n\right|r^{n-1}<\epsilon \end{equation}
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 $\epsilon$?.