# Infer the convergence of the coefficients of a power series based on the radius of convergence?

I'm taking a course in complex analysis and have been given a homework assignment. Part of the assignment includes True/False type questions, where we are given a mathematical statement and are required to show whether it is true or false. Here is one question from the assignment:

If the radius of convergence of a power series $$z\mapsto\sum_{n=0}^\infty a_nz^n$$ is $$2$$, then the series $$\sum_{n=0}^\infty |a_n|$$ converges.

Some of my classmates were struggling with this question and asked the professor for a hint. She (the professor) said that the statement was false, but obviously didn't provide a proof, as that's the homework problem. This however baffled me, since I was able to construct a very simple argument that the statement was true. Here's what I came up with:

If the series $$\begin{equation*} z\mapsto \sum ^{\infty }_{n=0} a_{n} z^{n} \end{equation*}$$ Has a radius of convergence of $$\displaystyle 2$$, then the series $$\begin{equation*} z\mapsto \sum ^{\infty }_{n=0} a_{n}( 2z)^{n} =\sum ^{\infty }_{n=0} 2^{n} a_{n} z^{n} \end{equation*}$$ Has a radius of convergence of $$\displaystyle 1$$. Having a look at the ratio test for this series, $$\begin{gather*} \lim _{n\rightarrow \infty }\left| \frac{2^{n+1} a_{n+1} z^{n+1}}{2^{n} a_{n} z^{n}}\right| =|2z|\lim _{n\rightarrow \infty }\left| \frac{a_{n+1}}{a_{n}}\right| < 1\\ \lim _{n\rightarrow \infty }\left| \frac{a_{n+1}}{a_{n}}\right| < \frac{1}{|2z|} \end{gather*}$$ Since the radius of convergence is $$\displaystyle 1$$, we can take e.g $$\displaystyle z=3/4$$, $$\begin{equation*} \lim _{n\rightarrow \infty }\left| \frac{a_{n+1}}{a_{n}}\right| < \frac{2}{3} < 1 \end{equation*}$$ Hence by the ratio test the sum of the $$\{\displaystyle a_{n}\}$$ converges absolutely, i.e, $$\sum_{n=0}^\infty |a_n|$$ converges.

Either my teacher misspoke, or there is an error in my argument, but I cannot find it. The only thing that I don't have a proof for is the assertion that if a series $$\sum_n a_nz^n$$ has a radius of convergence of $$R$$, then the series $$\sum_n a_n (\lambda z)^n$$ has a R.O.C of $$R/\lambda$$. It seems pretty obvious and I can't think of an instance where this would be false.

Can someone help please? Is my proof right?

• yes, you're right about all this. Commented Mar 26, 2021 at 13:57
• @DavidC.Ullrich Thanks for the feedback. I'll send my teacher an email. Commented Mar 26, 2021 at 13:58
• Note that the ratio test can only be used if all $a_n$ are non-zero. Commented Mar 26, 2021 at 14:07

In fact everything the OP said here was correct, including most of the proof. I have this pet peeve - it bugs me when people automatically turn to that $$|a_n|^{1/n}$$ thing any time the words "radius of convergence" appear. As is often the case we get a much simpler proof starting with the definition of "radius of convergence":
Saying that the radius of convergence is $$2$$ means the series converges for every $$z$$ with $$|z|<2$$ and diverges for every $$z$$ with $$|z|>2$$. In particular $$\sum a_n(3/2)^n$$ converges. So $$|a_n|(3/2)^n$$ is bounded, so $$|a_n|\le c(2/3)^n$$, so $$\sum|a_n|<\infty$$.
To my way of thinking this shows that the answer is obviously yes, as opposed to an argument that diddles with $$|a_n|^{1/n}$$...
• Thanks for the answer, but I don't quite understand $$\sum a_n(3/2)^n\text{ converges }\implies |a_n|(3/2)^n\text{ is bounded}$$ Can you elaborate? I'm probably missing something obvious. Commented Mar 26, 2021 at 14:10