Apparent paradox with power series convergence

Question

I'm having problem figuring out how these three facts are mutually consistent:

1. Call a set $S \subset \mathbb{C}$ to be concirclic if for some real number $R$, if you let $A$ be the (open) ball centered at zero with radios $R$ and $\overline{A}$ be the closure of $A$ then $A \subset S \subset \overline{A}$. For any power series $f(z) = \sum a_i z^i$, let $S_f := \{ z \in \mathbb{C} | f(z) \text{ is finite } \}$. Then $S$ is concirclic

2. If you let $f(z) = \sum (z-.5)^i$ then $S_f = \{ z \in \mathbb{C}, |z - .5| < 1 \}$. In particular, $S_f$ is not concirclic.

3. If you define $b_i$ as the sum of coefficients of $z^i$ in $\sum_{k \geq 0} (z-.5)^k$, then $b_i$ is well defined (i.e is not divergent). Then $f(z) = \sum b_i z^i$, and hence $S_f$ is concirclic.

I'm making some very stupid mistake but can't figure it lol

Answer 1

You defined $S_f$ as the domain of convergence of Taylor's series centered at $0$, and such domain is indeed concirclic (upd: if radius of convergence is finite). But in 2 you take Taylor series centered at $\frac{1}{2}$ - and it's domain of convergence in general differs from domain of convergence of series centered at $0$. So $S_f \neq \{z: |z - 0.5| < 1\}$

If you calculate $b_i$ you will note that $b_i = \frac{2}{3} \cdot \left(\frac{2z}{3}\right)^i$, and $\sum b_i z_i$ converges at, say $z = -1$, while original series diverges there.

Answer 2

The short answer is you cannot arbitrarily rearrange/regroup a series and always expect it to stay convergent/divergent. For example, $0+0+0+\dots$ converges, but $(1-1)+(1-1)+\dots=1-1+1-1+\dots$ diverges. You start with $\sum_{k\ge 0}(z-1/2)^k$, which is absolutely convergent on an open circle of radius $1$ centered at $1/2$. After some rearrangements, you end at $\sum_i b_i z^i$. This is still convergent on the original set, and equal to the original series there, but is also convergent on the larger circle which is centered at the origin with radius $3/2$.

In detail, you have $$ \sum_{k\ge 0}(z-1/2)^k=\frac1{1-(z-\frac12)}=\frac23\cdot \frac1{1-\frac23z}=\sum_{i\ge 0}\underbrace{(2/3)^{i+1}}_{b_i}z^i $$

The fact that your rearrangements lead to a correct answer in the original region of convergence (the circle of radius $1$ centered at $1/2$) suggests that they can be justified in this region, but for now I cannot see how.