# 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 $R>1$?  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 $z\in\mathbb{R}$, $1<z<\frac{3}{2}$, which contradicts the fact that the series diverges for all $|z|>1$. So we must have $R>1$ as desired.

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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.

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actually, do we need both sums to be absolutely convergent? – sigma Jul 11 '11 at 2:24
Not "both sums" but the "double sum". Otherwise you have the standard $a_{ij}=1$ if $j=i$, $-1$ if $j=i+1$, $0$ in all other cases example where $\sum_{i\ge 0}\sum_{j\ge 0}a_{ij}=0+0+0+\dots=0$ and $\sum_{j\ge 0}\sum_{i\ge 0}a_{ij}=1+0+0+\dots=1$ – fedja Jul 11 '11 at 13:13

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.

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But $1/(1+x)$ doesn't meet the positivity condition on the Taylor coefficients. – Gerry Myerson Jul 11 '11 at 2:47
True. I should have said "the statement minus that condition". Since the proof doesn't use that condition, it still follows that there must be something wrong in the proof. (As it turns out from fedja's answer, apparently what's wrong is just that one of the steps needs to be justified by invoking the positivity condition.) – joriki Jul 11 '11 at 3:22

I haven't read through your proof, but $\sum z^n/n^2$ is analytic, has positive real coefficients, and radius of convergence 1.

EDIT: Sorry, I was operating under the false impression that the dilogarithm is an entire function. I forgot that there's a branch cut along the real axis from 1 to infinity, which invalidates it as a counterexample here.

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