I want to show that $\sum_{k=0} ^n z^k $, $ \; |z| <1$, converges using the M-test.
So I want to choose $M_k = r^k$ for some $|z| \leq r <1$, where we want $r$ to be a fixed number.
If $z$ is fixed, I just fix the $r$ based on $z$.
But I think $z^k$ is a function of $z$ (for each $k$), so how can I fix $r$?
$\\$
From wiki
Suppose that {$f_n$} is a sequence of complex-valued functions defined on a $A$, and that there is a sequence of positive numbers $M_n$ satisfying $$\forall n \geq 1, \forall x \in A: \ |f_n(x)|\leq M_n,$$ $$\sum_{n=1}^{\infty} M_n < \infty$$ Then the series $\sum_{n=1}^{\infty} f_n (x)$ converges absolute and converges uniformly $A$.
This say that we want $M_k$ to an upper bound for $z^k$ for all $z$,but I don't understand how we can fix an $r$ unless we ignore the potential values of $z$ where $|z| > r$.
Is not so obvius using the M-Weierstrass test, but you can easly show it for $|z|<1-\delta$ taking $r \in (1-\delta ,1)$ then since is valid for all $1>\delta>0$ then aproach the limit
$M_n= r^n$