Uniform convergence of Taylor series I am trying to show:
If $f: B(0,R) \to \mathbb C$ is analytic then the Taylor series of $f$ at $0$ converges uniformly to $f$ in $B(0,r)$ for all $r\in (0,R)$
But I got stuck with my proof. Please can somebody help me? Here is what I have so far:
From here I have 
$$ f(z) = \sum_{n=0}^\infty z^n {1\over 2 \pi i}\int_C {f(w) \over w^{n+1}}dw$$
Therefore the Taylor series of $f$ at $0$ is
$T_0(z) = \sum_{n=0}^\infty z^n c_n$
where $c_n = {1\over 2 \pi i}\int_C {f(w) \over w^{n+1}}dw$.
My idea is to apply the Weierstrass $M$-test here for $f_n(x) = z^n c_n$. It holds that $|z^n| < r^n$ and $|c_n|\le {M \over r^n}$ and therefore $|f_n(z)|\le M$. The problem is how to show
$\sum_{n=0}^\infty M < \infty$? It is obviously not true (consider $R=1$). Where is my mistake?

You basically have the right ideas. Let instead of using a circle of radius $r$ for your contours, use a radius that is slightly bigger. That is, let $C = \{ z\in \mathbb{C}: |z| = r'\}$, where $r < r' < R$. 
With this, you'll have a convergent geometric series as the upper bound for $\sum |f_n(z)|$.
