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In Theorem of Baby Rudin 8.1, Rudin said that

If series $\sum_{n=0}^\infty c_nx^n$ converges for $|x|<R$

then $\sum_{n=0}^\infty c_nx^n$ converges uniformly on $[-R+\epsilon, R-\epsilon]$, no matter which $\epsilon > 0$ is chosen.

and I don't know why I can't use the open interval (-R, R) instead of $[-R+\epsilon, R-\epsilon]$

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  • $\begingroup$ Your question is not clear right now in my opinion. Are you asking how to conclude that the series converges uniformly on $(-R,R)$? Because that is not necessarily true. $\endgroup$ Apr 1 '20 at 12:37
  • $\begingroup$ I believe OP is asking why the theorem cannot be strengthened to say that the series converges uniformly on $(-R,R)$. $\endgroup$ Apr 1 '20 at 12:43
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$\sum x^{n}$ converges for $|x| <1$ but the convergence is not uniform on $(-1,1)$. Reason: $x^{n}$ does not tend to $0$ unifiormly on this interval.

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Consider the harmonic series $$1+z+z^2+\cdots$$ You know that this series converge if $|z|<1$ and it fail to converge if $z=1$.

The point is that you don't have necessarily a good behavior in the extremes.

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The difference is the word uniform and the placement of quantifiers. If $(f_n)$ is a sequence of functions on an interval $I$, we say $(f_n)$ converges to $f$ pointwise if: $$ \forall x \in I,\forall \eta > 0, \exists N\in \mathbb N, \forall n \in \mathbb{N}, n > N \implies |f_n(x) - f(x) | < \eta $$ We say $(f_n)$ converges to $f$ uniformly on $I$ if: $$ \forall \eta > 0, \exists N \in \mathbb{N},\forall x \in I,\forall n \in\mathbb{N}, n > N \implies |f_n(x) - f(x)| < \eta $$ In the first convergence, $N$ may depend on $x$ and $\eta$. But in the second, $N$ needs to work for all $x$ simultaneously (or “uniformly”).

As an example, consider the sequence of functions $f_n(x) = x^n$ on the interval $(-1,1)$. The pointwise limit of the sequence is $0$.

We claim that $(f_n)$ does not converge uniformly to $0$ on $(-1,1)$. To show this, let $\eta = \frac{1}{2}$. Given $N \in \mathbb{N}$, let $x$ be any number in the interval $(\eta^{1/(N+1)},1)$. Then $|f_{N+1}(x)| = |x^{N+1}| > \eta$.

However, for any $\epsilon > 0$ $(f_n)$ does converge to $0$ uniformly on $(1-\epsilon,1-\epsilon)$. Given $\eta > 0$, choose $N$ such that $(1-\epsilon)^N < \eta$. Then for any $x$ with $|x| < 1-\epsilon$, and any $n > N$, $$|x|^{n} < (1-\epsilon)^n < (1-\epsilon)^N < \eta$$

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