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If $f_n$ is a sequence of holomorphic functions on $D$ converging uniformly to $f$ on compact subsets of $D$, then the Cauchy integral formula implies that derivatives of all orders converge uniformly on compact sets to the respective derivatives of $f$.

My question is about whether a partial converse is true. Suppose $0 \in D$, and that the Taylor coefficients of the $f_n$ converged to the Taylor coefficients of $f$, i.e. if $f_n = \sum_{k=0}^\infty a_k^{(n)} z^k$ and $f = \sum_{k=0}^\infty a_k z^k$, then $\lim_{n\to\infty} a_k^{(n)} = a_k$. Can I conclude that $f_n$ converges uniformly on compact sets to $f$?

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Counterexample: $f_n(z) = n!z^n.$ Here we have $a_{k,n} \to 0$ for each $k,$ while $(f_n(z))$ diverges for each $z\ne 0.$

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  • $\begingroup$ Turns out I needed these functions to be Schlict so that I could apply the Bieberbach conjecture on the tail of the difference. $\endgroup$ – user369210 May 19 '17 at 21:36

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