TITLE: Rudin Theorem 3.44 explanation
QUESTION [5 upvotes]: This is from Rudin PMA, theorem 3.44: 

Theorem: Suppose the radius of convergence of $\sum c_n z^n $ is $1$, and suppose $c_0 \geq c_1 \geq c_2 \geq \cdots$, $\lim_{n\to\infty} c_n = 0$. Then $\sum c_n z^n$ converges at every point on the circle $\vert z \vert = 1$, except possibly at $z = 1$.
Proof: Put $a_n=z^n$, $b_n=c^n.$ The hypothesis of Theorem 3.42 (which is Dirichlet's Test) are then satisfied since $$|A_n|=\left|\sum_{m=0}^nz^m\right|=\left|\dfrac{1-z^{n+1}}{1-z}\right|\le\dfrac{2}{|1-z|}$$ if $|z|=1,z\ne1$.

Statement of theorem 3.42: 

Theorem 3.42 Suppose
  (a) the partial sums $A_n$ of $\sum a_n$ form a bounded sequence;
(b) $b_0\geqslant b_1\geqslant b_2\geqslant \dots;$
(c) $\lim_{n\to \infty} b_n=0.$
  Then $\sum a_nb_n$ converges.

I have these questions: Where do we use the fact that radius of convergence is $1$?  and  Is this attempt correct: $|1-z^{n+1}|\le2$ because, $|1-z^{n+1}|\le|1|+|z^{n+1}|=1+|z|^{n+1}=1+1=2$ ?

REPLY [5 votes]: If the theorem is exactly as you stated it then no, the fact that the radius of convergence is $1$ is not used in the proof. (The other hypotheses imply that the radius of convergence is at least $1$, and if  it's greater than $1$ then the conclusion is obvious.)