Let $f_k$ be a sequence of rational functions analytic in the discs $\{ |z| < 1 + \epsilon_k\}$ (with some $\epsilon_k > 0$), which converge to an analytic function $f$ in every point $|z| < 1$ (the convergence is uniform on compact subsets of $\mathbb{E}$). Furthermore, we assume that in $z=1$ every derivative $f^{(n)}_k(1)$ converges to some $a(n)$ as $k$ goes to $\infty$ and that $\sum_{n=0}^\infty a(n)(z-1)^n$ converges in some disc $\{ |z - 1| < \delta\}$.
My question is: does $f$ already have an analytic continuation around $z=1$ with the above taylor expansion? It seems to be difficult here that we do not know whether the convergence of the $f^{(n)}_k(1)$ is uniform in a certain sense.
Remark: A positive answer not using that $\sum_{n=0}^\infty a(n)(z-1)^n$ is locally analytic would be very gratifying in my case.