I'm trying to solve a problem regarding absolute convergence of a series of complex functions and finding it's derivative. Especially, I do not know how to find the derivative for all complex numbers, but only on every bounded subset of the complex numbers. Could some of you confirm whether my approach is correct and how to show it for all complex numbers. Thanks in advance :-)
Show that $$f(z)=\sum_{n=0}^{\infty} \frac{z^{2 n}}{(2 n) !}$$ is absolutely convergent for all values of $z$.
And prove that $$f'(z)=\sum_{n=0}^{\infty} \frac{z^{2 n+1}}{(2 n+1) !}$$ for all $z$.
So far I have only learned about pointwise and uniform convergence for series of functions. So if I want to show absolute convergence in this case, couldn't it both be absolute pointwise or absolute uniform convergence?
i) I'm assuming here that writing "for all values of $z$" is equivalent to showing absolute convergence on $\mathbb{C}$.
For pointwise absolute convergence, choose a $z_0\in\mathbb{C}$.
I can show that $\sum_{n=0}^{\infty} |\frac{z_0^{2 n}}{(2 n) !}|$ is convergent for any $z_0\in\mathbb{C}$ using the ratio test.
For uniform absolute convergence, I can only show convergence on any bounded subset and therefore not all of $\mathbb{C}$.
We can find a $M\in\mathbb{R}$ s.t. for any $z\in \{z\in\mathbb{C}: |z|<M \}=K$ we have that:
$|\frac{z^{2 n}}{(2 n) !}|\leq \frac{M^{2 n}}{(2 n) !}$.
Showing that $\sum_{n=0}^{\infty} \frac{M^{2 n}}{(2 n) !}$ is convergent and using Weierstrass's M-test gives us that $f(z)$ converges uniformly on $K$. So it's not uniform convergent on $\mathbb{C}$.
ii) If I have shown uniform convergence on $K$ and noting that $\{\frac{z^{2 n}}{(2 n) !} \}^{\infty}_{n=0}$ is a sequence of holomorphic functions. Then I can use the corollary:
Corollary 4.19. Let $G \subseteq \mathbb{C}$ be open. Let $f_{0}, f_{1}, \ldots$ be a sequence of holomorphic functions on $G$ and assume that the infinite series $$ s(z)=\sum_{k=0}^{\infty} f_{k}(z), \quad z \in G $$ converges locally uniformly on $G$ to a function $s: G \rightarrow \mathbb{C} .$ Then $s \in \mathcal{H}(G)$ and the series can be differentiated term by term infinitely often $$ s^{(p)}(z)=\sum_{k=0}^{\infty} f_{k}^{(p)}(z), \quad z \in G, p=1,2, \ldots $$ and each of these series converges locally uniformly on $G$.
To show that $f'(z)$ is given by the above series by differentiating $f(z)$ term by term and rewriting the series.
However, I have only shown this is true for all $z\in K$ (So for every bounded subset of $\mathbb{C}$).
How do I argue that this is true for all complex numbers?