I try to prove:
If $f$ is analytic on an open disk $B(0, R) \subseteq \mathbb C$ where $R>0$ then the radius of convergence of its Taylor series is $\ge R$.
But I stuck with my proof. Please can someone help me? Here is what I have so far:
Let $f:B(0,R) \to \mathbb C$ be analytic on $B(0,R)$ and let $T_0(x) = \sum_{n = 0}^\infty c_n z^n$ denote its Taylor series at $0$. The radius of convergence is given by
$$ r = {1 \over \lim \sup \sqrt[n]{|c_n|}}$$
where
$$ c_n = {1 \over 2 \pi i} \int_\gamma {f(w) \over w^{n+1}}dw$$
where $\gamma$ is a counterclockwise curve around $0$ and contained in $B(0,R)$.
Now how to show $r \ge R$? If $\sqrt[n]{|c_n|} < {1\over R}$ it would be useful but how? Because $f(w)$ can be arbitrary.
Best Answer
To finish on the way you started, let $\gamma$ be a circle with centre $0$ and radius $0 < \rho < R$. Then the standard estimate tells you
$$\lvert c_n\rvert = \frac{1}{2\pi}\left\lvert\int_{\lvert w\rvert = \rho} \frac{f(w)}{w^{n+1}}\,dw\right\rvert \leqslant \frac{1}{2\pi} \int_0^{2\pi} \frac{\lvert f(\rho e^{i\varphi})\rvert}{\rho^n}\,d\varphi \leqslant \rho^{-n}\max \{\lvert f(\zeta)\rvert : \lvert\zeta\rvert = \rho\}.$$
That gives you an upper bound on $\limsup\limits_{n\to\infty} \sqrt[n]{\lvert c_n\rvert}$ in terms of $\rho$. As $0 < \rho < R$ was arbitrary, you get the desired conclusion.
Another way to obtain the result is to use Cauchy's integral formula,
$$f(z) = \int_{\lvert w\rvert = \rho} \frac{f(w)}{w-z}\,dw,$$
for $\lvert z\rvert < \rho$, and expand $\dfrac{1}{w-z}$ into a geometric series, which you know converges uniformly on the circle $\lvert w\rvert= \rho$. Interchanging summation and integration then yields the convergence of the Taylor series in $\lvert z\rvert < \rho$. Let $\rho\to R$.