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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.

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    $\begingroup$ Take $h(z) = e^{1/z},f_k(z) = h(z-1-\frac{1}{ k})$ then $\forall n,\lim_{k \to \infty} f_k^{(n)}(1) = 0$. $\endgroup$
    – reuns
    Dec 12, 2017 at 11:06
  • $\begingroup$ Thanks, got the same counter-example meanwhile... Fortunately I can exclude any essential singularities in my case: I know that the $f_k$ are (without loss of generality) rational functions, in particular meromorphic in the entire plane. $\endgroup$ Dec 12, 2017 at 11:30
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    $\begingroup$ $f_k(z ) = \sum_{m=0}^{e^{e^k}} \frac{1}{m!} (\frac{1}{z-1-1/k})^m$ $\endgroup$
    – reuns
    Dec 12, 2017 at 12:01
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    $\begingroup$ Given that the Runge theorem allows you to approximate anything analytic in $\{|z|<1-3\delta\}\cup\{|z-1|<2\delta\}$ uniformly by polynomials on $\{|z|\le 1-4\delta\}\cup\{|z-1|\le \delta\}$ with any precision you want, asking the question in this form is a rather poor idea. Why not to tell the real setup without an attempt to generalize? $\endgroup$
    – fedja
    Dec 13, 2017 at 1:06

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