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).

  • 1
    $\begingroup$ 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 $x=1$. $\endgroup$
    – rogerl
    Commented Jun 24, 2019 at 20:55
  • $\begingroup$ 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? $\endgroup$
    – Dominic
    Commented Jun 24, 2019 at 21:01
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    $\begingroup$ @Dominic no, because absolute convergence implies convergence. $\endgroup$ Commented Jun 24, 2019 at 21:02
  • $\begingroup$ 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. $\endgroup$
    – Dominic
    Commented Jun 24, 2019 at 21:10

2 Answers 2


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 $0<r<\infty$).


I'd like to provide a hopefully clear and concise answer. First, let's rephrase the question: can a power series be conditionally convergent on an interval $(a-R,a+R)$ but only absolutely convergent on a smaller interval $(a-r,a+r)$ where $r < R$? Since the ratio or root tests are used to prove the big theorem that's often implicitly used in the background for finding radii of convergence, the answer is "No." The statement of that theorem is usually along the following lines:

From Stewart's Calculus, 9th Edition

Basically, radius of convergence only makes sense in the context of absolute convergence (there isn't a separate notion for conditional convergence of a power series). So using the radii mentioned in the rephrasing I posted above, the series must diverge outside the interval $[a-r, a+r]$, and therefore is divergent on at least some of the interval $(a-R,a+R)$. Conditional convergence is very much a boundary phenomenon: it can only occur on the boundaries of finite intervals in the context of power series. There may be some cleanup necessary in my above explanation to be very precise, but that's the basic idea.


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