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).
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
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 $0<r<\infty$).