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Here is one argument that I understand for why this is true, using the comparison test.

Suppose the series $$\sum_{n=1}^{\infty}c_nx^n$$ converges for $|x|<R$.

We want to show that the series converges absolutely in the interval $[-R+\epsilon, R-\epsilon]$

Choose $\zeta $ such that $R-\epsilon<R-\zeta<R$

Then for $x\in[-R+\epsilon, R-\epsilon]$,

$$|c_nx^n|\leq |c_n (R-\epsilon)^n|=|c_n||R-\zeta|^n\frac{|R-\epsilon|^n}{|R-\zeta|^n}.$$ But since $\sum c_n(R-\epsilon)^n$ converges, there exists $M$ such that $M\geq |c_n(R-\epsilon)|^n$ for all $n$.

This, $$|c_nx^n|\leq M\frac{|R-\epsilon|^n}{|R-\zeta|^n},$$ and the result follows by comparison with the geometric series.

My real question is, in Rudin, it is claimed that every power series converges absolutely within the radius of converges using the root test.

I understand the formula given by $$R=\frac{1}{\limsup_{n\to\infty} |c_n|^{\frac{1}{n}}}.$$

I know that the series $\sum c_n (R-\epsilon)^n$ converges for any epsilon positive, but I cannot make any conclusion about the value of $$\limsup_{n\to\infty} |c_n(R-\epsilon)|^{\frac{1}{n}}.$$ (Since the converse to root test may not be true).

Thanks in advance

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2 Answers 2

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  1. Given a series $f(x) = \sum_n c_n (x-x_0)^n$, we define the "radius of convergence" as $R = \frac1{\limsup |c_n|^{1/n}}$. With no proof of the wanted property of $R$, this is just a name.

  2. The root test says that $\sum_n a_n $ converges absolutely if $\limsup |a_n|^{1/n} < 1$.

  3. Applied to our series $f(x)$, we get absolute convergence at $x$ if $$ \limsup_{n\to\infty} |c_n|^{1/n} |x-x_0| < 1.$$ It is easy to check that this is true if $|x-x_0|<R$.
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$\lim \sup |c_n (R-\epsilon)^{n}|^{1/n}=\frac {R-\epsilon} R$. Since $\frac {R-\epsilon} R<1$ the conclusion follows.

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  • $\begingroup$ How did you figure out that equality? $\endgroup$ Feb 15, 2020 at 5:37
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    $\begingroup$ @Jungleshrimp $|(R-\epsilon)^{n}|^{1/n}=R-\epsilon$ and $\lim\sup c|a_n|=c \lim \sup |a_n|$ for any constant $c \geq 0$. $\endgroup$ Feb 15, 2020 at 5:48

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