Power series with finite radius of convergence but analytical continuation to a strictly bigger disc? Is there a complex power series $\sum_{n=0}^\infty a_n z^n$, with a finite radius of convergence $0<r<\infty$, that admits an analytic continuation to a disc $\{z\in \mathbb{C} : |z|<R\}$ with strictly bigger radius $R>r$? (The analytic continuation should be defined everywhere on the bigger disc.)
 A: No.
Suppose $f(z)=\sum_{n=0}^\infty a_n z^n$ whenever $|z|<r$,
and also that $f(z)$ is holomorphic on a bigger disc $|z|<R$.
It is a theorem that if $f(z)$ is holomorphic on the disc $|z|<R$,
there is a Taylor series $\sum_{n=0}^\infty b_n z^n$ that is convergent on $|z|<R$ and that is equal to $f(z)$.
Now, $a_n=\dfrac{f^{(n)}(0)}{n!}=b_n$.
A: No: suppose there is some holomorphic function $f$ defined on some open disk with radius $r'>r$ which agrees with $\sum a_n z^n$ on $|z| < r$. Then by the Taylor expansions for analytic functions, $f$ equals its Taylor series $\sum c_n z^n$ on $|z| < r'$ (here $c_n = f^{(n)}(0) /n!$).
But by assumption $f = \sum c_n z^n$ agrees with $\sum a_n z^n$ on $|z| < r$. It follows that $a_n = c_n$ for all $n$ (otherwise if $a_n \neq c_n$ for some $n$, they would have different $n'$th derivatives at zero).
But now using the formula for the radius of convergence, we get the contradiction
$$
r = ({\lim \sup}_{n \to \infty} \sqrt[n]{|a_n|})^{-1} = ({\lim \sup}_{n \to \infty} \sqrt[n]{|c_n|})^{-1} \geq r'.
$$
