Problem in understanding Cauchy - Hadamard theorem. I am now studying power series in Complex Analysis.There I found a theorem for determining radius of convergence of a power series known as "Cauchy - Hadamard theorem" which states that "For a given power series $$\sum_{i=0}^{\infty} a_{n}(z-a)^{n}$$,define the number $R$, $0\leq R \leq \infty$, by $\frac 1 R = \lim \sup |a_{n}|^{\frac1 n}$.If $0<R<\infty$ then 
(a) if $|z-a|< R$, the series converges absolutely;
(b) if $|z-a| > R$ the series become unbounded and so the series diverges;
(c) if $0 < r < R$, then the series converges uniformly on $\{z : |z-a| \leq r\}$."
But I dont find any difference between (a) and (c).I think in any compact subset of $\{z :|z-a| < R\}$ the power series is absolutely and unifomly convergent.So there is no difference between absolute and uniform convergence inside $\{z :|z-a| < R\}$.Is it true?Please help me anyone in understanding this concept.
Thank you in advance.
 A: It is true that in a compact subset of $\{z :|z-a| < R\}$ the power series converges absolutely and uniformly, but neither (a) nor (c) say both of those things.  Rather, (a) points out the series is absolutely convergent at every point in the open disk and (c) points out that the series converges uniformly on the smaller closed disks (equivalent to converging uniformly on compact subsets).  In general, convergence can be uniform but not absolute (even for a power series converging on a closed disk), or absolute (pointwise) but not uniform, so neither asserts that the other is true.  When combined they give your conclusion that the series converges absolutely and uniformly on compact subsets.
A: Example: the geometric series $\sum_{n=0}^{\infty}z^n$ converges absolutely for $|z|<1$ and diverges for $|z| \ge 1$. Hence $R=1$. Furthermore
$\sum_{n=0}^{\infty}z^n= \frac{1}{1-z} $ for $|z|<1$ .
For $|z|<1$ we have 
$|\sum_{k=0}^{n}z^n- \frac{1}{1-z}| =\frac{|z|^{n+1}}{|1-z|}$, hence
$|\sum_{k=0}^{n}z^n- \frac{1}{1-z}| \to \infty$ for $z \to1$.
This shows that  $\sum_{n=0}^{\infty}z^n$ does not converge uniformly on $\{z :|z| < 1\}$.
If $0<r<1$ and $|z| \le r$, then $|z|^n \le r^n$. Since $\sum_{n=0}^{\infty}r^n$ is convergent, the Weierstrass M - test shows that $\sum_{n=0}^{\infty}z^n$  converges uniformly on $\{z :|z| \le r\}$.
