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

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  • $\begingroup$ In the proof I think you mean $b_n = c^n$. If I remember correctly assuming the radius of convergence is $1$ is only a normalisation — the goal of this theorem is to prove convergence of this series exactly on the boundary of the disc where this series converges. The last derivation in your question looks correct to me. $\endgroup$ Feb 11 '18 at 13:20
  • $\begingroup$ Ok, thank you. Making correction $\endgroup$
    – Silent
    Feb 11 '18 at 13:24
  • $\begingroup$ @TheoreticalEconomist, so can we get rid of that assumption and theorem still holds? $\endgroup$
    – Silent
    Feb 11 '18 at 13:27
  • $\begingroup$ actually, now that I think about it, we can’t. The bound shown in the proof works only when $\vert z \vert \le 1$. I misremembered. :) $\endgroup$ Feb 11 '18 at 13:30
  • $\begingroup$ I am sorry, but I can't understand! Can you please elaborate? @TheoreticalEconomist $\endgroup$
    – Silent
    Feb 11 '18 at 13:37
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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.)

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  • $\begingroup$ What hypothesis does imply that radius of convergence is at least $1$? $\endgroup$
    – Silent
    Feb 11 '18 at 14:33
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    $\begingroup$ Saying the $c_n$ decrease to $0$ implies that $|c_n|$ is bounded, hence $\sum c_n z^n$ converges for $|z|<1$. $\endgroup$ Feb 11 '18 at 14:35
  • $\begingroup$ Ok, thanks for that comment. I now see that $\sum c_n z^n$ converges for $|z|<1$,. But still can't think why radius of convergence is at least $1$. Rudin defines radius of convergence as $\frac{1}{\limsup_{n\to\infty}\sqrt[n]{|c_n|}}$ $\endgroup$
    – Silent
    Feb 11 '18 at 14:44
  • $\begingroup$ I don't have the book here. II find it very hard to believe that that's how he defines the radius of convergence. Surely the definition is something else, and then that formula iis a theorem? The right definition is this: $R$ is the radius of convergence if the series converges whenever $|z|<R$ and diverges whenever $|z|>R$. $\endgroup$ Feb 11 '18 at 14:47
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    $\begingroup$ The fact that $r\ge1$ iis also obvious from that formula: Since $c_n\to0$ we have $|c_n|\le1$ for $n>N$... $\endgroup$ Feb 11 '18 at 15:02
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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.

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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 $R < 1$.

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  • $\begingroup$ I am thankful for this elaborate reply. I just can't understand why $\frac{R^{n+1}}{\vert 1 - z \vert} \le \left\vert\frac{ 1 - z^{n+1} }{1-z}\right\vert + \left\vert\frac{ 1 }{1-z}\right\vert$? $\endgroup$
    – Silent
    Feb 11 '18 at 14:28
  • $\begingroup$ @Silent This is because $R^{n+1} = \vert z^{n+1} \vert \le \vert z^{n+1} - 1 \vert + \vert 1 \vert$. $\endgroup$ Feb 11 '18 at 14:30
  • $\begingroup$ Why is $R^{n+1} = \vert z^{n+1} \vert$? (I am feeling stupid, but can't figure this out.) $\endgroup$
    – Silent
    Feb 11 '18 at 14:31
  • $\begingroup$ @Silent Recall that by assumption $R = \vert z \vert$, so $R^{n+1} = \vert z \vert^{n+1}= \vert z^{n+1} \vert$. $\endgroup$ Feb 11 '18 at 14:34

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