Rudin Theorem 3.44 explanation

Question

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, $|1-z^{n+1}|\le|1|+|z^{n+1}|=1+|z|^{n+1}=1+1=2$ ?

Answer 1

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.)

Answer 2

Well, the theorem would not make sense, or be needed, without the hypothesis that the radius is $1$. If the radius is less than that, you are assured by definition that the series diverges when $|z| = 1$. If it is greater than that, the series converges (uniformly) for all $|z| = 1$. So $r=1$ is the only interesting case.

Answer 3

Suppose that the radius of convergence of the power series $\sum c_n z^n$ is $R$, where $R$ is not necessarily equal to $1$.

We know by definition of $R$ that the power series converges whenever $\vert z \vert <R$, and diverges whenever $\vert z \vert > R$. We wish to investigate the convergence of this series for $\vert z \vert = R$ using Dirichlet's test (theorem 3.42). Hence, assume from now that $\vert z \vert = R$.

Dirichlet's test is conclusive only if $R \le 1$. If $R > 1$, the partial sums $A_n$ are unbounded. To see this, observe that

$$ \frac{R^{n+1}}{\vert 1 - z \vert} \le \left\vert\frac{ 1 - z^{n+1} }{1-z}\right\vert + \left\vert\frac{ 1 }{1-z}\right\vert = \left\vert A_n \right\vert + \left\vert\frac{ 1 }{1-z}\right\vert . $$

This means that $\left\vert A_n \right\vert$ grows without bound when $R > 1$. (This, of course, doesn't mean that the series diverges for $\vert z \vert = R > 1$; we just need a test other than Dirichlet's to say anything about it.)

Notice now that Rudin's proof applies verbatim for $R \le 1$. In particular, we still have the bound

$$ \left\vert A_n \right\vert \le \frac{2}{\vert 1 - z \vert} $$

when $\vert z \vert = R < 1$. In other words, the assumption that $R = 1$ isn't of particular importance; what is really important for what (in my opinion) Rudin is attempting to prove is that $R \le 1$.

Of course, if you view this theorem as just an investigation of the convergence of the series for $\vert z \vert = 1$, then the assumption that $R = 1$ is (as noted in the other answers) not needed. However, the theorem is trivial whenever $R > 1$, and false whenever $R < 1$.