Do power series converge uniformly and absolutely on radius of convergence? 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 $0 ≤ |x − c| &lt; R$ and diverges for
$|x − c| &gt; R$. Furthermore, if $0 ≤ \rho &lt; R$, 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 $&lt;1$ to conclude the series is $&lt; \infty$. 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 $0 ≤ |x − c| &lt; R$ and diverges for
$|x − c| &gt; R$.

Is there a reason to introduce the number $\rho$ which I'm missing? Or can the text actually be simplified like this? Thank you!
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 $|x|&lt;R$ and that it diverges when $|x|&gt;R$ is $R=1$. However, the series $(1)$ does not converge uniformly when $|x|&lt;1$. 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 $\rho&lt;R$ and if that proof uses the Weierstrass $M$-test, then you can deduce from that proof that the series $(2)$ converges absolutely when $|x-c|&lt;R$.
Fixing $\rho &lt; R$ allows you to use
$(\rho/R)^n \to 0$.