There is a theorem in complex analysis stated as "if a complex power series (centered at z_0) converges absolutely at z = z_1, then it converges uniformly in region: |z-z_0| <= |z_1-z_0|."
I know how to prove this theorem, but I wonder what will happen if we loosen the restriction in the statement – absolute convergence – to just convergence? Does that hold, as well?
If not, is there any counter example? Or if you could kindly help me to understand this statement better, I really appreciate that.
The statement just reminds me of the difference of "conditional convergence" and "absolute convergence" in real analysis, so quite naively, I bring this idea to complex analysis. Sorry if it's a dumb question.
Best Answer
It is not true. For instance, the series $\sum_{n=1}^\infty\frac{(-1)^n}n$ converges, but the series $\sum_{n=1}^\infty\frac{z^n}n$ does not converge uniformly when $|z|\leqslant1$. Actually, it diverges when $z=1$.
But the absolute convergence still holds. In fact, it is enough to assume that the sequence $\left(a_n(z_1-z_0)^n\right)_{n\in\Bbb N}$ is bounded. If $B$ is such that $(\forall n\in\Bbb Z_+):\left|a_n(z_1-z_0)^n\right|\leqslant B$ and if $|z-z_0|<|z_1-z_0|$, then, if $n\in\Bbb N$,\begin{align}\sqrt[n]{\left|a_n(z-z_0)^n\right|}&=\sqrt[n]{\left|a_n(z_1-z_0)^n\left(\frac{z-z_0}{z_1-z_0}\right)^n\right|}\\&\leqslant\sqrt[n]B\left|\frac{z-z_0}{z_1-z_0}\right|\end{align}and so, since $\lim_{n\to\infty}\sqrt[n]B=1$, if you take some $c\in\left(\left|\frac{z-z_0}{z_1-z_0}\right|,1\right)$, you have $\sqrt[n]{\left|a_n(z-z_0)^n\right|}\leqslant c$ if $n$ is large enough. So, by the root test, your series converges absolutely.