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 statement is correct and the proof is essentially correct though the fact that all coefficients are positive is crucial. You actually used it without noticing: when you talked about exchanging the order of summations, you were a bit sloppy because we need the terms in the double series to be summable in absolute value to use "sequential Fubini". Fortunately, the originally inner sum consists of terms of the same sign, which allows to say that the absolute value of that sum is the same as the sum of absolute values and reduce the property that we really need to the one you declared as sufficient.
