Power series: If $|z-z_0|< R$ the series converges absolutely I'm trying to prove absolute convergence of the power seris $$\sum_{n=0}^{\infty} a_n (z-z_0)^n, \qquad |z-z_0| < R$$ where $R^{-1} = \limsup |a_n|^{1/n}$.
WLOG, suppose $z_0=0$ (otherwise consider $w=z-z_0$). If $|z|<R$ there exists an $r$ such that $|z| < r < R$. Hence $$\frac{1}{R} < \frac{1}{r},$$ so that by the definition of $R$ there exists an $N \in \mathbb{N}$ such that for all $n \geq N$, $$|a_n|^{1/n} < \frac{1}{r},$$ or $$|a_n| < r^{-n}.$$ Hence $$|a_n z^n| < \left( \frac{|z|}{r} \right)^n, \quad \forall n \geq N.$$ Therefore, $$\sum_{n \geq N} |a_n z^n| < \sum_{n \geq N} \left( \frac{|z|}{r} \right)^n.$$ Here's my problem: I want to show convergence of the last series. I know that $|z|<r$ by the choice of $r$, but this is not enough to show that the last series converges, am I right?
Hope you can help.
 A: The question was answered in Daniel Fischer's comment: the series $ \sum_{n \geq N} \left( \frac{|z|}{r} \right)^n $ is geometric, with ratio $|z|/r<1$. Thus it converges.
We even know its sum:
$$ \sum_{n \geq N} \left( \frac{|z|}{r} \right)^n = \frac{|z|^{N}/r^N}{1-|z|/r} < \frac{1}{1-|z|/r} \tag{1}$$
I'll squeeze out a little extra out of (1), under the additional assumption that there exists $N$ such that  $|a_n|^{1/n}\le R^{-1}$ for all $n\ge N$. In this case we can use $r=R$ and keep $N$ the same for all $z$ in the disk. Then (1) implies the growth estimate
$$|f(z)|\le \frac{M}{R-|z|},\quad |z|<R\tag{2}$$
The assumption $|a_n|\le R^{-n}$ can be replaced with $|a_n|R^n $ being bounded, because $f$ can be multiplied by an appropriate constant to make  $|a_n|R^n \le 1$. Thus, we get a nice corollary: 

If $f(z)=\sum_{n=0}^\infty a_n z^n$ where the coefficients are bounded, then $(1-|z|)\,|f(z)| $ is bounded in the unit disk.

This sort of boundedness property comes up in complex analysis: see Bloch space
