# Power series with finite radius of convergence but analytical continuation to a strictly bigger disc?

Is there a complex power series $$\sum_{n=0}^\infty a_n z^n$$, with a finite radius of convergence $$0, that admits an analytic continuation to a disc $$\{z\in \mathbb{C} : |z| with strictly bigger radius $$R>r$$? (The analytic continuation should be defined everywhere on the bigger disc.)

• Given a function $f(z)$, when talking about radius of convergence one has to specify the point $z_0$ around which this convergence takes place. Oftentimes, this is assumed that $z_0 = 0$ since if not, one looks at function $g(z) = f(z+z_0)$. But this is important distinction since, for example, for function $f(z) = \frac{1}{1+z}$ radius of convergence at $0$ is $1$ but at $100$ (i.e. $100 + 0)i$ it is $99$. Commented Dec 23, 2022 at 16:50
• Since I wrote sum a_n z^n I assumed that it is implicitly clear that I was talking about a series of powers of z, hence the center of the disk of convergence is z=0 Commented Dec 23, 2022 at 18:56
• I was not being critical about your exposition. Just making a point that at different location radius of convergent may be different. Commented Dec 23, 2022 at 22:05

No: suppose there is some holomorphic function $$f$$ defined on some open disk with radius $$r'>r$$ which agrees with $$\sum a_n z^n$$ on $$|z| < r$$. Then by the Taylor expansions for analytic functions, $$f$$ equals its Taylor series $$\sum c_n z^n$$ on $$|z| < r'$$ (here $$c_n = f^{(n)}(0) /n!$$).

But by assumption $$f = \sum c_n z^n$$ agrees with $$\sum a_n z^n$$ on $$|z| < r$$. It follows that $$a_n = c_n$$ for all $$n$$ (otherwise if $$a_n \neq c_n$$ for some $$n$$, they would have different $$n'$$th derivatives at zero).

But now using the formula for the radius of convergence, we get the contradiction $$r = ({\lim \sup}_{n \to \infty} \sqrt[n]{|a_n|})^{-1} = ({\lim \sup}_{n \to \infty} \sqrt[n]{|c_n|})^{-1} \geq r'.$$

• Sorry, but I don't understand why that answers my question. This just says that the given power series does not converge outside of |z|<r. Why could there not be other power series (not centered at zero) that converge outside of the disc |z|<r and whose values agree with the given one inside |z|<r, and these cover a disc strictly bigger? Commented Dec 23, 2022 at 13:24
• @MichaelBächtold I edited my answer, but I'm not sure this is enough.. I'm thinking about it for a bit longer. Commented Dec 23, 2022 at 14:00
• @MichaelBächtold I think now I have a valid proof. It uses Taylor's theorem for holomorphic functions, but this can be found in many books. Commented Dec 23, 2022 at 14:23
• Right, silly me. Should have just looked up that theorem Commented Dec 23, 2022 at 19:02

No.

Suppose $$f(z)=\sum_{n=0}^\infty a_n z^n$$ whenever $$|z|,

and also that $$f(z)$$ is holomorphic on a bigger disc $$|z|.

It is a theorem that if $$f(z)$$ is holomorphic on the disc $$|z|,

there is a Taylor series $$\sum_{n=0}^\infty b_n z^n$$ that is convergent on $$|z| and that is equal to $$f(z)$$.

Now, $$a_n=\dfrac{f^{(n)}(0)}{n!}=b_n$$.

• See my comment to the other answer. I would have the same question with your answer. Commented Dec 23, 2022 at 13:25
• @MichaelBächtold the answer has been edited for clarity Commented Dec 23, 2022 at 15:08