Power series Convergence, same Coefficients, different Powers

Question

If $g(z) = \sum_{n \in \mathbb{N}} a_n \cdot (z-z_0)^n$ converges absolutely, how can I formally prove that $f(z) = \sum_{n \in \mathbb{N}} a_n \cdot (z-z_0)^{n-1}$ also converges absolutely?

I know that the $\frac{d}{dz} g(z) = \sum_{n \in \mathbb{N}} n \cdot a_n \cdot (z-z_0)^{n-1}$ has the same radius of convergence as $g$ and multiplying each summand of $f$ with a natural number will only increase the absolute value of $\frac{d}{dz} g$ compared to $f$.

How to prove without writing prose? Ty!

Answer 1

For $z\neq z_0$ you have $f(z)=\frac{g(z)}{z-z_0}$. Multiplying an absolutely convergent series by a complex number (like $\frac1{z-z_0}$) does not change its absolute convergence.

Answer 2

Assume that $\sum_{n} a_n (z-z_0)^n$ converges absolutely for some $z$, i.e., $\sum_n |a_n| \rho^n$ is convergent, with $\rho := |z-z_0|$. Clearly, we can assume $\rho > 0$ (otherwise the statement is trivial).

But then also $\sum_n c_n$, with $c_n := |a_n| \rho^{n-1} \geq 0$ is convergent, since $c_n = (|a_n| \rho^n) / \rho$.