Prove the Uniform Convergence of a Certain Series on Any Compact Subset of C $D \subset C$ is a connected open set. $a \in D$, $f \in H(D)$, the series $\sum_{n=0}^\infty f^{(n)}(a)$ converges. Prove that:


*

*$f$ can be extended to an analytic function on $C$.

*$\sum_{n=0}^\infty f^{(n)}(z)$ converges uniformly on every compact subset of $C$.
(I am stuck on question 2. After achieving this series $\sum_{n=0}^\infty f^{(n)}(z)=\sum_{n=0}^\infty \sum_{m=0}^\infty f^{(m+n)}{(a)}\frac{(z-a)^m}{m!}$, I wonder if a limit interchange can be justified.)
 A: Let us denote $a_n = f^{(n)}(a)$ and $S_m = \sum_{n=0}^m a_n$.
In a neighborhood of $z=a$ the function $f$ is equal to its Taylor series:
$$
\tag{*}
 f(z) = \sum_{n=0}^\infty a_n \frac{(z-a)^n}{n!} \, .
$$
The convergence of $(S_m)$ implies that $a_n \to 0$, and in particular that $(a_n)$ is a bounded sequence. It follows that the power series in $(*)$ has infinite radius of convergence, and therefore defines an entire function $F$.
$F$ is an analytic continuation of $f$ because the functions coincide in a neighborhood of $z=a$ (or because they have the same Taylor series).

For the second part we need the derivatives of $F$, which are
$$
 F^{(k)}(z) = \sum_{n=0}^\infty a_{n+k} \frac{(z-a)^n}{n!} \, .
$$
Now let $K \subset \Bbb C$ be a compact set and $\epsilon > 0$. Set $M = \max \{ e^{|z-a|} \mid z \in K \}$.
$(S_m)$ is a Cauchy sequence, so that $|S_p - S_q| < \epsilon$ for sufficiently large $q > p \ge N$. For $z \in K$ and $p > q \ge N$ we then have
$$
 \sum_{k=p+1}^q F^{(k)}(z) = \sum_{n=0}^\infty (S_{n+q}- S_{n+p}) \frac{(z-a)^n}{n!} $$
and therefore
$$
 \left| \sum_{k=p+1}^q F^{(k)}(z) \right| \le \epsilon \sum_{n=0}^\infty \frac{|z-a|^n}{n!} = \epsilon e^{|z-a|} \le M \epsilon \, .
$$
It follows that $ \sum_{k=0}^\infty F^{(k)}(z)$ is uniformly convergent on $K$.
