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This is according to Tao Theorem 15.1.6 (c) the last part. Let $R > 0$ be the radius of convergence of some formal power series $$ \sum_{n=0}^\infty c_n(c-a)^n $$ Then we may define $$ f: (a-R,a+R) \rightarrow \mathbb R: x \mapsto \sum_{n=0}^\infty c_n(x-a)^n $$

We know that for all $0 < r < R$ the series $\sum_{n=0}^\infty c_n(x-a)^n$ converges uniformly on $[a-r,a+r]$ (Weierstrass M-test) to a continuous function $f$ which is the function $f$ defined above restricted to that compact interval. Now I have to show that $f$ is continuous. My attempt:

Let $x_0 \in (a-R,a+R)$. Then $|a-x_0| < R$. Then we can find some $r \in \mathbb R$ such that $$ |x_0-a| < r < R $$ In particular $|x_0-a| \leq r < R$. Thus $x_0 \in [a-r,a+r]$ with $0 < r < R$. On that interval $f$ is continuous. Thus $f$ is in $x_0$ continuous. Since $x_0$ was arbitrary the claim follows.

Is that correct ?

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You should state $|x_0-a|<r$ implies $x_0\in(a-r,a+r)$. $f$ restricted to $[a-r, a+r]$ is continuous at the interior point $x_0$; so $f$ is continuous at the point $x_0$. –  David Mitra Jan 19 '13 at 16:14
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The idea is correct, but your write up is a bit off. Knowing that $x_0\in [a-r, a+r]$ tells you that $f$ restricted to $[a-r,a+r]$ is continuous. Then $f$ will be continuous at $x_0$ if $x_0\in (a-r,a+r)$ (you can't conclude the same from the foregoing if $x_0$ is one of $a-r$ or $a+r$).

But you're ok; in fact $|a-x_0|<r$ implies $x_0\in(a-r,a+r)$. So change your second to last paragraph to something like:

"We have $|x_0-a| <r < R$. Thus $x_0 \in (a-r,a+r)\subset[a+r,a-r]$ with $0 < r < R$. On the interval $ (a-r,a+r)$, $f$ is continuous. Thus $f$ is continuous at the point $x_0$. Since $x_0$ was arbitrary the claim follows."

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Thanks for your answer. Just for clarification of your first paragraph: You say we know $f$ is continuous on $[a-r,a+r]$ from which we may not conclude that that if $x_0 \in \{a-r,a+r\}$ follows that $f$ is continuous in $x_0$ ? –  André Jan 19 '13 at 16:37
    
@André Yes. You could only conclude the appropriate one-sided continuity. –  David Mitra Jan 19 '13 at 16:47
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