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I am reading the following passage in my lecture notes on power series (and have seen similar statements in other texts):

Let $$R=\sup \left\{\left|x\right|\ge 0:\sum a_nx^n \text{ converges}\right\}.$$ If $R=0$, then the series converges only for $x=0$. If $R>0$, then the series converges absolutely for every $x\in \mathbb R$ with $|x|<R$, since it converges for some $x_0\in\mathbb R$ with $|x|<|x_0|<R$.

I'm confused about $|x_0|<R$.

If $M=\sup (A)$, then for every $M'<M$, there exists an $x\in A$ such that $x>M'$. Then, using the definition of an upper bound (in e.g. Spivak's Calculus), it follows that $M'<x\color{red}{\leq} M$. Suppose $A=\{1,2,3\}$, then $2.8<3$, but stating $2.8<3<3$ would be incorrect.

Why is it correct then to state $|x_0|<R$ in this case?

psie
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  • Think about a series which diverges on some, or even all, points on the boundary of its disk of convergence. (Useful rule of thumb: since $<$ is stronger than $\le$, a statement with a "$<$-hypothesis" is weaker than a statement with an "$\le$-hypothesis.") – Noah Schweber Aug 14 '23 at 20:18
  • Hmm, such a series would be $\sum x^n$. However, I do not understand why there is $x_0\in \mathbb R$ such that $|x|<|x_0|<R$. Where does this inequality come from? – psie Aug 14 '23 at 20:44
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    To be fair, it might just have been a typo from your teacher. But yes, it should be $|x| < |x_0| \leq R$, since, like you mentioned, when a $\sup$ is a $\max$, there is no guarantee that there will exist elements that get closer and closer to the $\sup$ without being equal to it, finite sets are the perfect example for this. – Bruno B Aug 14 '23 at 21:43
  • @BrunoB Thank you for replying. I suspect it is a typo too, but I have found it in other textbooks too, which is strange (see here for instance). – psie Aug 14 '23 at 21:48
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    Seems like it's a bad use of the definition of $R$ as $\sup S$ in that case. There is an element between $|x|$ and $R$ but it doesn't have to be different from $R$. – Bruno B Aug 14 '23 at 21:53

1 Answers1

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By the definition of the supremum, we have $|x| \lt |x_0| \leq R$. Now, $\sum a_nx^n$ converges for $x=x_0$ and so for all $x\in\mathbb R$ with $|x|<|x_0|$. We can find an $|x_1|$ such that $|x|<|x_1|<|x_0|\leq R$ and we have obtained the desired inequality.

psie
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