I don't know how to solve this: Fix a positive real number $k$ and set $p=1/k$. For $n \ge 0$, consider the function $f_n : \mathbb{C} \rightarrow \mathbb{C}$ defined by $f_n(z)=k^n z^n$.
Fix $z_0 \in \mathbb{C}$ with $\mid z_0\mid \ge p$. Show that the numerical series $\sum_{n\ge0}f_n(z_0)$ diverges.
To show that the series diverges, we show that $ f_n(z_0)$ does not converge to $0$ as $n \rightarrow \infty$. This common test for divergence of real series also works for series of complex numbers.
To show this, simply notice that for any $n$, $$ |f_n(z_0)| = \left|k^n z_0^n\right| = k^n \; \left|z_0 \right|^n. $$ As $\left|z_0 \right| \geq p = 1/k$, we must have that $k \left|z_0 \right| \geq 1$, and so $ k^n \; \left|z_0 \right|^n$ must be at least $1$, too. Thus $f_n(z_0)$ always is at least distance $1$ away from $0$, and so cannot converge to it.