Regarding the radius of convergence and its equality to a certain limit

Question

Let $f$ be a holomorphic function on the open unit disk $\mathbb{D}$, and suppose that $f$ cannot be extended holomorphically to any open set $\Omega$ containing $\overline{\mathbb{D}}$. Let $f(z) = \sum_{n=0}^\infty a_n z^n$ be the power series expansion of $f$ around $0$, and assume that $a_n \ne 0$ for all but finitely many $n$. Show that if the limit $\lim_{n \to \infty} |a_n / a_{n+1}|$ exists, then it is equal to $1$.

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Is this supposed to be obvious? It pretty much tells that the radius of convergence around $0$ is $R=1$. So of course within the disc series converges and outside the disc it diverges, whatever happens on the boundary depends on the function hence the convergence is not guaranteed.

Also what is up with assumption that $a_n \ne 0$ for all but finitely many $n$? Why doesn't it ensure that the limit in question exists?

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Edit. Alright, I looked at some literature (including proofs) and figured out that the limit $\lim_{n \to \infty} |a_n / a_{n+1}|$ is indeed nothing more than the radius of convergence itself. However I am still lacking an elegant argument...

Ideally I wanted to get from $\lim_{n \to \infty} |a_n / a_{n+1}|=L$ general case to $L \ge 1$ and $L \le 1$ by using the definition $R = \sup \{ r \ge 0 \mid |a_n| r^n \text{ is bounded} \}$. It would involve $|a_n| L^n$ and whatnot. I am stuck with this at the moment.

Answer 1

The quotient criterion says: $$\limsup\frac{|a_{n+1}z^{n+1}|}{|a_{n+1}z^{n+1}|}<1\implies\sum_n^\infty a_n z^n\text{ converges}$$ $$\liminf\frac{|a_{n+1}z^{n+1}|}{|a_{n}z^{n}|}>1\implies\sum_n^\infty a_n z^n\text{ diverges}$$ (Note the differences in $\limsup$ and $\liminf$ due to different approaches...)

As the limit exists that boils down to: $$|z|<\lim\frac{|a_n|}{|a_{n+1}|}\iff\sum_n^\infty a_n z^n\text{ converges}$$ That is the radius of convergence is given precisely by: $$R=\lim\frac{|a_n|}{|a_{n+1}|}$$

Answer 2

Due to holomorphicity the power series converges within the unit disk $\mathbb{D}$, however it is impossible to find any disk of bigger radius (and centered at the origin) with the same property of convergence. (In conclusion the radius of convergence of the power series in question is exactly $R = 1$.)

Now let us assume the existence of the limit $\lim_{n \to \infty} |a_n / a_{n+1}| = L$.

If $r<L$, then $|a_k / a_{k+1}| > r$ eventually. Consequently $|a_k| > r|a_{k+1}|$ for all $k \ge N$ and \begin{equation*} |a_N| r^N \ge |a_{N+1}| r^{N+1} \ge |a_{N+2}| r^{N+2} \ge |a_{N+3}| r^{N+3} \ge \ldots. \end{equation*} Hence the sequence $|a_k|r^k$ is bounded and $r \le 1$ because one is the supremum involved in such sequences. Since $r<L$ is arbitrary we also have $L \le 1$.

If $s>L$, then $|a_k / a_{k+1}| < s$ eventually. Consequently $|a_k| < s|a_{k+1}|$ for all $k \ge N$ and \begin{equation*} |a_N| s^N \le |a_{N+1}| s^{N+1} \le |a_{N+2}| s^{N+2} \le |a_{N+3}| s^{N+3} \le \ldots. \end{equation*} Hence the terms $a_k z^k$ do not converge to $0$ for $|z| \ge s$ (and the whole power series does not converge) so $s \ge 1$. Since $s>L$ is arbitrary we also have $L \ge 1$.

It is then an obvious inequality $1 \le L \le 1$ that gives us \begin{equation*} \lim_{n \to \infty} \left| \frac{a_n }{ a_{n+1}} \right| = 1. \end{equation*}