An example of non uniform convergence.

Question

$(a_n)$ is a positive decreasing sequence that converges to $0$, we claim that $\sum a_n z^n$ converges uniformly on $U=\{z \in \mathbb C, |z| \lt 1 \text{ and } |z-1| \lt 1\}$. I don't think this is true and I suspect the condition $|z-1| \lt 1$ is the issue, but I can't prove it, I see that if $z \to 1$ and $a_n = {1 \over n}$ for example the series "blows off to infinity" (?). The series converges pointwise, so summation by parts on the rest translates to: $$|R_n(z)|=|\sum_{k=n+1}^{\infty}\frac{1-z^{k+1}}{1-z}(a_k-a_{k+1})|$$ I wasn't able to either prove or disprove the claim using this, what did I miss?

Answer 1

You are right. On $U$, that power series converges to $-\log(1-z)$ (here, $\log$ is the main branch of the logarithm). But that function is unbounded on $U$ ($\lim_{z\to1}\bigl\lvert-\log(1-z)\bigr\rvert=\infty$) and therefore it cannot be the uniform limit of a sequence of functions each of which is bounded on $U$.