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Let $\sum_{k=0}^{\infty}a_k(x-a)^k$ be a power series with real coefficients $a_k$, the constant $a \in \mathbb{R}$ and the positive radius of convergence $R>0$. Let \begin{equation} D = \begin{cases} ]a-R, \ a+R[ & \text{if } R < \infty \\ \mathbb{R} & \text{if } R = \infty \ \end{cases} \end{equation} and $ \ f: D \rightarrow \mathbb{R}$, $ \ f(x) := \sum_{k=0}^{\infty}a_k(x-a)^k$. $ \ f$ is then continuous on $D$.

This is from my lecture notes. I am not really sure, what is happening here. Do I understand this correctly? A power series has a certain radius of convergence $R$. Depending on $R$ the set $D$ is defined. $D$ is the set where the power series converges (?). $D$ is then taken as the domain for a function which is defined as the power series. $f$ is then continuous on $D$.

Maybe somebody can explain it more clearly or add a helpful image.

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  • $\begingroup$ Yes, you are absolutely right. I forgot the coefficients in the sum. $\endgroup$ – Jennifer W. Sep 8 '17 at 12:49
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This follows from uniform convergence. The series does not have uniform convergence on $D$, but for every compact subset of $D$ it has. This means that $f$ is continuous on every compact subset of $D$ and therefore everywhere in $D$.

Note that the radius of convergence doesn't say what happens on the endpoints. For $|x-a|=R$ it may converge, but it may also fail to converge (and it may differ between the end points). So $D$ may in fact be a set slightly smaller than the set where $f$ converges.

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