Do power series converge uniformly and absolutely on radius of convergence?

Question

I was reading this text on power series. On pages $182-183$, the author gives a proof of the following statement:

Let $$ \sum_{n=0}^{\infty} a_n (x-c)^n $$ be a power series. There is a non-negative, extended real number $0 \le R \le \infty$ such that the series converges absolutely for for $0 ≤ |x − c| < R$ and diverges for $|x − c| > R$. Furthermore, if $0 ≤ \rho < R$, then the power series converges uniformly on the interval $|x − c| ≤ \rho$.

During the proof, the author first proves that the series converges absolutely by bounding $|a_nx^n|$ using inequalities leading to a geometric series, and then using that the common ratio is $<1$ to conclude the series is $< \infty$. After this, the author then proves that the series converges uniformly on $|x − c| \le \rho$ by again using inequalities that lead to a geometric series, but this time he uses the Weierstrass M-test to conclude that the series converges uniformly.

This left me a bit confused because it is my understanding that from the M-test you can conclude that the series is both uniformly and absolutely convergent. And since the statement of the theorem allows for $|x − c| = \rho$, I don't understand why the author separates the statement (and the proof) into two parts when I believe you could simplify it to

Let $$ \sum_{n=0}^{\infty} a_n (x-c)^n $$ be a power series. There is a non-negative, extended real number $0 \le R \le \infty$ such that the series converges uniformly and absolutely for $0 ≤ |x − c| < R$ and diverges for $|x − c| > R$.

Is there a reason to introduce the number $\rho$ which I'm missing? Or can the text actually be simplified like this? Thank you!

Answer 1

This final statement of yours is false. Take, for instance, the geometric series$$\sum_{n=0}^\infty x^n.\tag1$$The only $R$ for which it is both true that the series $(1)$ converges absolutely when $|x|<R$ and that it diverges when $|x|>R$ is $R=1$. However, the series $(1)$ does not converge uniformly when $|x|<1$. On the other hand, for each $\rho\in(0,1)$ the series $(1)$ does converge uniformly when $|x|\leqslant\rho$.

However, it is true that if you prove that a power series$$\sum_{n=0}^\infty a_n(x-c)^n\tag2$$converges uniformly when $|x-c|\leqslant\rho$, for each $\rho<R$ and if that proof uses the Weierstrass $M$-test, then you can deduce from that proof that the series $(2)$ converges absolutely when $|x-c|<R$.

Answer 2

Fixing $\rho < R$ allows you to use $(\rho/R)^n \to 0$.

Answer 3

In the first blockqoute $\rho < R$ is need in case the series does not converge at $x = c \pm R$. That is $$ \sum_{n=0}^\infty a_n (\pm R)^n = \pm \infty $$ then the convergence on $|x-c|<R$ is not uniform.

Added: if the series converges on $(c-R,c+R)$ but not at $x = c \pm R$, then there is (actually for all) $\varepsilon >0$ such that the partial sums differ from the limit function on $(c-R,c+R)$ by more than that $\varepsilon$, since the limit function is unbounded near $x = c \pm R$, in contrast to any partial sum.