Let $D = \{z \in \mathbb C: |z|< 1\}$ and let $A(D)$ be the disk algebra, i.e the Banach algebra of functions $D \to \mathbb C$ that are analytic in $D$ and continuous in $\overline D$ with the supremum norm in D, denoted by $\| \cdot \|$. Define $$ S_{m,n} = \left \{ f \in A(D): \sum_{k=0}^m \frac{\|f^{(k)}\|}{k!} > n\right \} \text{ and } S_n = \bigcup_{m \geq n} S_{m,n}. $$
I'm trying to proof that each $S_n$ is an open subset of $A(D)$.
My attempt: Given $n \geq 1$ and $f \in S_n, \, \exists m\geq n, \, f \in S_{m,n}$, i.e $$ \sum_{k=0}^m \frac{\|f^{(k)}\|}{k!} > n. $$
So, I have to find a $\delta > 0$ such that if $g \in A(D)$ and $\|f-g\| < \delta$, $g \in S_n$, i.e $g \in S_{M,n}$ for some $M\geq n$.
$$ \sum_{k=0}^M \frac{\|g^{(k)}\|}{k!} \geq \sum_{k=0}^M \frac{\|f^{(k)}\|}{k!} - \sum_{k=0}^M \frac{\|g^{(k)} - f^{(k)}\|}{k!} > n - \sum_{k=0}^M \frac{\|g^{(k)} - f^{(k)}\|}{k!}. $$
However, I don't see how I can conclude the proof.
Help?
First, there's a typo in your definition of $S_{m,n}$: surely the index on the sum should be $k$, not $n$. And the notation $\|f^{(k)}\|$ seems unfortunate, since $f^{(k)}$ need not be an element of the disk algebra. If we write the definition like so
$$S_{m,n} = \left \{ f \in A(D): \sum_{k=0}^m \frac{\sup_{|z|<1}|f^{(k)}(z)|}{k!} > n\right \}$$
then it's easy to show that $S_{m,n}$ is open. Two hints:
(i) $f\in S_{m,n}$ if and only if $f\in A(D)$ and there exist $z_0,\dots,z_m\in D$ with $$\sum_{k=0}^m\frac{|f^{(k)}(z_k)|}{k!}>n,$$
(ii) If $f\in A(D)$ and $z\in D$ then $$|f^{(k)}(z)|\le\frac{k!||f||}{(1-|z|)^k}$$(apply "Cauchy's Estimates" in $D(z,1-|z|)$).