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

Question

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?

Answer 1

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 $|a_n|^{1/n}$...