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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 $R>1$? $$ $$ 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? $$ $$ 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|>1$. So we must have $R>1$ as desired.

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