Uniform convergence of $\sum_{n=0}^{\infty} \frac{(-1)^n 2i}{(2n+1)z^{2n+1}}$ on squarewith vertices $\pm6\pm6i$?

Question

Can someone explain me how I can check whether the convergence of $\sum_{n=0}^{\infty} \frac{(-1)^n 2i}{(2n+1)z^{2n+1}}$ is uniform on the (boundary of) square $A$ with vertices $\pm6\pm6i$?

Answer 1

When we have a series

$$\sum_{n=0}^\infty \frac{(-1)^n 2i}{(2n+1)z^{2n+1}},$$

we can bound the size of the terms from above when we can bound the absolute modulus of $z$ from below.

On the square (boundary/contour) with vertices $\pm 6 \pm 6i$, we know that $\lvert z\rvert \geqslant 6$. Thus we can majorise the series by

$$\sum_{n=0}^\infty \left\lvert \frac{(-1)^n 2i}{(2n+1)z^{2n+1}} \right\rvert \leqslant \sum_{n=0}^\infty \frac{2}{(2n+1)6^{2n+1}} < \frac13\sum_{n=0}^\infty \frac{1}{36^n}$$

and see that the series converges uniformly on the contour. More, we can see that it converges uniformly on $\{ z \in\mathbb{c} : \lvert z\rvert \geqslant r\}$ for every $r > 1$ since the analogous computation yields a majorisation by a geometric series with ratio $r^{-2} < 1$.

Answer 2

This is intended as addition to @DanielFischer's answer, which covers the given question. In general, when one considers a Laurent series $$\sum_{n\in\mathbb{Z}}a_n(z-z_0)^n,$$ it's well-known (good reference by @Argon here) that the series converges in annulus $\{z\in \mathbb{C}\mid R_1<|z−z_0|<R_2\}$ for some $0\leq R_1,R_2$ and diverges outside the closure of the annulus (naturally, the series might diverge on the entire plane). Convergence on the boundary of the open annulus--or a subset of it--is possible, but not of interest here. As an aside, one can find $R_1,R_2$ based on knowledge of the $a_n$'s, as demonstrated on Wiki.

Following DanielFischer's arguments, you can see that the series converges uniformly on any subannulus, i.e., $$\{z\in \mathbb{C}\mid R^\prime_1\leq|z−z_0|\leq R^\prime_2\},$$ for $R_1<R^\prime_1\leq R^\prime_2<R_2$. In fact, as such closed annuli are compact (at least if $R^\prime_2<\infty$), and any compact subset is contained in such an annulus, it's customary to say that Laurent series converge uniformly on compacta. Here, the given square is a compact subset of the region of convergence (namely, $\{z\in\mathbb{C}\mid |z|>1\}$, as DanielFischer also showed).