# On the definition of radius of convergence; a small supremum technicality

• schniefen
schniefen
Homework Statement
I have a question about a definition of the radius of convergence, in particular about a small detail regarding a property of the supremum.
Relevant Equations
If ##M=\sup (A)##, then for every ##M'<M##, there exists an ##x\in A## such that ##x>M'##.
I am reading the following passage in these lecture notes (chapter 10, in the proof of theorem 10.3) 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##.

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?

You are right, it should be ##|x|<|x_0|\le R##.

schniefen
Your definition of sup is not complete:
##M=\sup A\Longleftrightarrow##
1) ##x\in A\Rightarrow x\le M##
2) for any ##M'<M## it follows that there exists ##x\in A## such that ##x>M'##

schniefen
I thought maybe the statement ##|x|<|x_0|< R## is not incorrect after all if the set $$\left\{\left|x\right|\ge 0:\sum a_nx^n \text{ converges}\right\}$$ is an interval, but this is not something we know a priori. Besides I am really not sure how to show the set is an interval -- probably requires a definition of what an interval is. Anyway, this seems far-fetched and I lean towards a typo, though I have seen this stated in other texts as well (see for instance A First Course in Mathematical Analysis by Brannan, chapter 8, section 8.3).

schniefen said:
Why is it correct then to state ##|x_0|<R## in this case?
You can go directly from the statement for ##|x_0| \le R## to ##\exists x_1 : |x| \lt |x_1| \lt |x_0| \le R## to ##|x| \lt |x_1| \lt R##.

schniefen
FactChecker said:
You can go directly from the statement for ##|x_0| \le R## to ##\exists x_1 : |x| \lt |x_1| \lt |x_0| \le R## to ##|x| \lt |x_1| \lt R##.
Not directly, but by proving that the series is convergent at ##x_1##. That is not a problem, just a redundant step.

upd

schniefen said:
Besides I am really not sure how to show the set is an interval
Theorem. If the series is convergent at ##x'## then it is convergent in the interval ##\{|x|<|x'|\}##. (It is an interval in ##\mathbb{R}## and a circle if we consider ##\mathbb{C}##)

This theorem is proved in any textbook where power series are discussed. Find it.

Last edited:
PeroK, FactChecker and schniefen
Ok, I think this clarified it. Thanks a lot!

To summarize; 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|##. As @FactChecker pointed out, we can find an ##|x_1|## such that ##|x|<|x_1|<|x_0|## and we are done.

schniefen said:
Ok, I think this clarified it. Thanks a lot!

To summarize; 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|##. As @FactChecker pointed out, we can find an ##|x_1|## such that ##|x|<|x_1|<|x_0|## and we are done.
Yes. But as @wrobel pointed out above, there is no point in doing this. Using ##\le## as you originally suggested is more directly derived from the definitions.

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