A series of functions $\sum_{n=0}^{\infty}\frac{a_n}{z^n}$ is given. The exercise is to prove that the series $\sum_{n=0}^{\infty}\frac{a_n}{z^n}$ converges locally uniformly on $\mathbb{C}\backslash \overline{K(0,1)}$ and uniformly on $\overline{K(2+2i,1)}$ given that the series of numbers $\sum_{n=0}^{\infty}a_n$ converges.
Remark: $K(z,r)$ is an open circle around $z$ of radius $r$ and $\overline{K(z,r)}$ is a closed circle around $z$ of radius $r$.
I know that when I prove that the given series of functions converges locally uniformly on $\mathbb{C}\backslash \overline{K(0,1)}$, I can easily prove that it also converges uniformly on $\overline{K(2+2i,1)}$ since $\overline{K(2+2i,1)}$ is a compact set and local uniform convergence implies uniform convergence on compact subsets.
However, I don't know whether I should try to prove the local uniform convergence by definition or by some other known theorem/result. It seems very intuitive that it does indeed converge locally uniformly to $f(z)=0$ by definition, but I don't know how to write out the formal proof. I therefore tried writing that for every $z\in \mathbb{C}\backslash \overline{K(0,1)}$ there exists $r>0$ such that $K(z,r)\subseteq \mathbb{C}\backslash \overline{K(0,1)} $ (which is obvious) and that $f|_{K(z,r)} \rightrightarrows 0$ (which I don't know how to prove).
Let $K$ be a compact subset of $\Bbb C\setminus\overline{K(0,1)}$ and let $m=\min\{|z|\mid z\in K\}$. Then $m>1$. Since the series $\sum_{n=0}^\infty a_n$ converges, $\lim_{n\to\infty}a_n=0$, and therefore the sequence $(a_n)_{n\in\Bbb N}$ is bounded. So, by the comparison test $\sum_{n=0}^\infty\frac{|a_n|}{m^n}$ converges. So, since$$(\forall z\in K):\left|\frac{a_n}{z^n}\right|\leqslant\frac{|a_n|}{m^n},$$your series converges uniformly on $K$.