Radius of convergence of the given power series.

Question

Let $f$ be a holomorphic function in the open unit disc such that $\lim_{z\to 1}f(z)$ doesn't exist. Let $\sum_{n=0}^\infty a_n z^n$ be the taylor series of $f$ about $z=0$ and $R$ be the radius of convergence. Then

(a) $R=0$

(b)$0<R<1$

(c) $R=1$

(d) $R>1$

My idea: $R\leq1$. Since $\lim_{z\to 1}f(z)$ doesn't exist. So, $\sum_{n=0}^\infty a_n z^n$ doesn't converges at $z=1$. We know that $\frac{1}{1-z}$ has a radius of convergence $1$. So, $R=1$. But I don't know How to prove it exactly. Please help me.

Answer 1

The answer is that $R=1$:

• If $R>1$, then the series $\sum_{n=0}^\infty a_nz^n$ converges to a holomprphic function on the open disk centered at $0$ with radus $R$. Therefore, $\lim_{z\to1}f(z)$ exists.
• You can't have $R<1$ because the Taylor series of a holomorphic function converges at every point of every open disk contained in its domain. And you know that it converges at the open disk centered at $0$ with radius $1$.