Taylor Series expansion at z=0, and radius of convergence I have the following question: Consider the domain
$$
D=B(0,1)\cup B\left(\frac{1}{2}, 1\right)
$$
It is given
that $f:D\rightarrow \mathbb{C}$ is an analytic function in $D$, and $f^{(n)}(0)$ is a positive real number for every positive integer $n$. Let $R$ be the radius of convergence of the Taylor series of $f$ at $z=0$. Is it true that $R>1$?
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$$
I have attached my proof to the following problem, although I am not sure if it correct, as I have clearly not used the fact that $f^{(n)}(0)$ is a positive real number for every positive integer $n$. How do I make use of this fact to prove/disprove the statement?
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$$
Proof:
Since $f$ is analytic on the ball $B(0,1)$, it follows from the definition of radius of convergence that $R\geq1$.
Suppose on the contrary that $R=1$.
By Taylor's Theorem, we may express $f$ as a Taylor series at $z=0$ as follows:
$$f(z)=\sum_{n=0}^\infty\frac{f^{(n)}(0)}{n!}z^n$$
where the series converges absolutely for all $z\in B(0,1)$, and diverges for all $|z|>1$.
Thus, by differentiating both sides of the above equation $k$ times, we have that for all $z\in B(0,1)$,
$$
f^{(k)}(z)=\sum_{n=k}^\infty\frac{f^{(n)}(0)}{(n-k)!}z^{n-k}.
$$
Also, since $f$ is analytic on the ball $B\left(\frac{1}{2},1\right)$, it follows from Taylor's Theorem that we may also express $f$ as a Taylor series at $z=\frac{1}{2}$ as follows:
$$
f(z)=\sum_{k=0}^{\infty}\frac{f^{(k)}\left(\frac{1}{2}\right)}{k!}\left(z-\frac{1}{2}\right)^k,
$$
where the series converges absolutely for all $z\in B\left(\frac{1}{2},1\right)$.
Now, by setting $z=\frac{1}{2}$, we have that for all $k\geq0$,
$$
f^{(k)}\left(\frac{1}{2}\right)=\sum_{n=k}^{\infty}\frac{f^{(n)}(0)}{(n-k)!}\cdot\frac{1}{2^{n-k}}.
$$
Then for all $z\in B\left(\frac{1}{2},1\right)$, we have the following:
$$
f(z)
=\sum_{k=0}^{\infty}\frac{f^{(k)}\left(\frac{1}{2}\right)}{k!}\left(z-\frac{1}{2}\right)^k
=\sum_{k=0}^{\infty}\sum_{n=k}^{\infty}\frac{f^{(n)}(0)}{(n-k)!k!}\cdot\frac{1}{2^{n-k}}\cdot\left(z-\frac{1}{2}\right)^k
$$
$$
=\sum_{n=0}^{\infty}\sum_{k=0}^n\frac{f^{(n)}(0)}{(n-k)!k!}\cdot\left(\frac{1}{2}\right)^{n-k}\cdot\left(z-\frac{1}{2}\right)^k
=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}\sum_{k=0}^n\frac{n!}{(n-k)!k!}\cdot\left(\frac{1}{2}\right)^{n-k}\cdot\left(z-\frac{1}{2}\right)^k
$$
$$
=\sum_{n=0}^{\infty}\frac{f^{(n)}(0)}{n!}z^n.
$$
Note: The interchanging of the summations is possible as the series $\sum_{k=0}^{\infty}\frac{f^{(k)}\left(\frac{1}{2}\right)}{k!}\left(z-\frac{1}{2}\right)^k$ converges absolutely for all $z\in B\left(\frac{1}{2},1\right)$; this follows from the Rearrangement Theorem, where any rearrangement of an absolutely convergent series converges to the same sum as the original series.
This implies that the Taylor series of $f$ at $z=0$ converges for all $z\in B\left(\frac{1}{2},1\right)$; and in particular for all $z\in\mathbb{R}$, $1<z<\frac{3}{2}$, which contradicts the fact that the series diverges for all $|z|>1$. So we must have $R>1$ as desired.

The first sentence in your proof is wrong, though its conclusion is correct. While it's true that the radius of convergence extends to the nearest point where the function is not holomorphic, this is a theorem and doesn't follow directly from the definition of the radius of convergence.
A counterexample to the statement is given by the function $1/(1+x)$, so there must be something wrong somewhere in your proof.
