If I have that $\sum_{n=0}^{+\infty} a_n(z-z_0)^n$ converges and $\sum_{n=0}^{+\infty} \frac{b_n}{(z-z_0)^n}$ converges, then obviously $\sum_{n=0}^{+\infty} \left(a_n(z-z_0)^n+\frac{b_n}{(z-z_0)^n}\right)$ converges, where $z,z_0,a_n,b_n \in \mathbb{C}$ and $z \neq z_0$.
Does the converse hold?
Namely, if we have that $\sum_{n=0}^{+\infty} \left(a_n(z-z_0)^n+\frac{b_n}{(z-z_0)^n}\right)$ converges, than it always follows that $\sum_{n=0}^{+\infty} a_n(z-z_0)^n$ converges and $\sum_{n=0}^{+\infty} \frac{b_n}{(z-z_0)^n}$ converges?
I really think that it is not true, but I'm stuck in looking for a counterexample. Any hint would be appreciated.
Thank you!
EDIT: I know that in the general case (generic complex series) if we have that $\sum_{n=0}^{+\infty} (z_n+w_n)$ converges, then it is not necessary true that $\sum_{n=0}^{+\infty} z_n$ or $\sum_{n=0}^{+\infty} w_n$ converge.
As an example: $\sum_{n=0}^{+\infty} (n+0i)$ diverges, $\sum_{n=0}^{+\infty} (-n+0i)$ diverges, but $\sum_{n=0}^{+\infty} [(n+0i)+(-n+0i)]$ converges to $0$.
Given the fact that $z-z_0=c$ a constant, let $a_n=\frac{1}{c^n}$ and $b_n=-c^n$, both series diverge, but the series of pairwise sums $=0$. I don't think this is what you had in mind.