I want to show that for all $\delta > 0$, the power series $\sum_{n=0}^\infty a_n z^n$ converges uniformly on the set $S:=\{z \in \mathbb C \mid |z-1|\geq \delta,|z|\leq1\}$ if $a_n$ is a monotonically decreasing series of positive, real numbers (edit: and $\lim_{n \to \infty} a_n = 0$).
My idea was to apply the weierstrass criterion: If I can show $\sum_{n \in \mathbb N} \lVert a_n z^n \rVert_\infty < \infty$, the power series converges uniformly (right?).
But when I try to prove it, I end up with $\lVert a_n z^n \rVert_\infty = a_n$,
$$\lVert a_n z^n \rVert_\infty=\sup_{z \in S} |a_n z^n|=a_n \sup_{z \in S} |z|^n=a_n$$
and that doesn't imply $\sum_{n \in \mathbb N} \lVert a_n z^n \rVert_\infty < \infty$ because e.g. $\sum_{n \in \mathbb N} \frac 1n = \infty$.
Is my attempt correct? Did I make a mistake?
You cannot prove it, since it is false. Suppose that $(\forall n\in\mathbb Z_+):a_n=1$. Then your series is $\sum_{n=0}^\infty z^n$, which diverges when $\lvert z\rvert=1$. Or, if you dislike the fact that this sequence is constant, take $a_n=1+\frac1{n+1}$.
However, the statement is true if you add the hypothesis that $\lim_{n\to\infty}a_n=0$. Then it follows from Abel's test.