Suppose that $0 < a_{n+1} < a_n$ for all $n$ and $a_n \rightarrow 0$. Let $r > 0$. Show that $\sum_{n=0}^{\infty} a_nz^n$ converges uniformly on the set where $|z| \leq 1$ and $|z-1| \geq r$.
My thought process:
1) I understand $|z| \leq 1$ is all the points in the unit circle. I do not know what $|z-1| \geq r$ is, nor do I know what the intersection of those looks like. Basically, what does the set that this converges on look like?
2) This looks sort of like a geometric series, but the $a_n$ term is throwing me off. Typically this series would converge to $\frac{a}{1-z}$ if $a_n$ was just a constant, right? How does that change when $a_n$ is not constant, but converging to zero?
3) I think I understand the definition of uniform convergence for a function, but how does that connect to a series? In other words, for $\varepsilon >0$, I need to show that $|f_n(z) - f(z)| < \varepsilon$ for all $n>N$, but what is $f_n$ and $f$? I would assume $f_n$ is the partial sum and $f(z)=\frac{a}{1-z}$. In other words, I need to show $|\frac{a(1-z^n)}{1-z}-\frac{a}{1-z} |< \varepsilon$? If that is correct I'm again stuck because I don't know how the $a_n$ changes this series.