Question on Taylor's theorem

Question

Taylor's Theorem

Let $\{c_n\}$ be a sequence of complex numbers.

Let $R$ be the radius of convergence of $\sum c_n z^n$.

Let $|b|<R$ and $f(x)=\sum c_n z^n$ on the disk $B(0,R)$.

Then, $\forall |x-b|<R-|b|, f(x)=\sum d_n (x-b)^n$ for some sequence $\{d_n\}$.

In the situation illustrated above, is $R-|b|$ the raius of convergence of $\sum d_n z^n$?

Answer 1

It can be larger. Take for instance $f(z) = \frac{1}{1+z^2}$. The radius of convergence at $0$ is $1$. The radius of convergence at $b=3/4$ is $5/4$, larger than $1/4$. Note that $5/4$ is the distance form $3/4$ to the set $\{i, -i\}$, the singular points of $f$.