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Is the radius of convergence of a conditionally convergent series always equal to the radius where it converges absolutely? For example, the power series:

$$ \sum_1^{\infty} (-1)^{n+1}\frac{2^n x^n}{n 3^n} $$ is absolutely convergent when $|x|<3/2$. What is the radius of convergence where the series is only conditionally convergent? Or are they they same?

I'm having some problems with a power series with coefficients having alternate signs and I can't explain why the Root Test is converging (numerically) to a value slightly higher than what I believe the convergence radius should be (0.85 vs. 0.86).

Dominic
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    The series $\sum_{0}^\infty (-1)^n\frac{x^n}{n}$ is absolutely convergent for $-1<x<1$, but is conditionally convergent (and not absolutely convergent) only for $x=1$. – rogerl Jun 24 '19 at 20:55
  • Ok thanks. I wonder if someone has an example where the domain of conditional convergence is a region, of some measure, less than the region of absolute convergence? – Dominic Jun 24 '19 at 21:01
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    @Dominic no, because absolute convergence implies convergence. – user10354138 Jun 24 '19 at 21:02
  • Ok thank you. I will look elsewhere for the discrepancy. I suppose my numeric data then is surely not yet precise enough to converge to the expected value of approx 0.8507. This one is peculiar because all the others (about 20) are converging nicely. – Dominic Jun 24 '19 at 21:10

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If the power series $$ \sum_{n=0}^{\infty}a_nx^n $$ converges absolutely for $x=b>0$, then it converges absolutely for $x\in[-b,b]$, by an easy comparison.

Conversely, if the series diverges for $x=c>0$, then it diverges for $|x|\ge c$, again by comparison.

If $r$ is the supremum of the set of $b\ge0$ such that the series converges absolutely for $x=b$, then it is easy to prove that

  1. the series converges absolutely for every $x$ with $|x|<r$;
  2. the series diverges for every $x$ with $|x|>r$.

This is why $r$ is called the radius of convergence. The special case when $r=0$ means that the series only converges for $x=0$; if $r=\infty$, then the series converges absolutely for every $x$.

At $r$ and $-r$ the series may converge (absolutely or conditionally) or diverge. The set of points where the series converges conditionally is a subset of $\{r,-r\}$ (provided, of course $0<r<\infty$).

egreg
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