Suppose $$|x-x_{0}|\leq\frac{1}{ \limsup\sqrt[n]{|a_n|}}$$ for the power series

$\sum\limits_{n=0}^{\infty} a_{n} (x-x_{0})^n$

My question is this, say if $x_a= x_{0}-\frac{1}{ \limsup\sqrt[n]{|a_n|}}$ is absolutely convergent, then can $x_b= x_{0}+\frac{1}{\limsup\sqrt[n]{|a_n|}}$ be divergent or conditionally convergent?


2 Answers 2


Let $ r= \frac{1}{\lim \sup\sqrt[n]{|a_n|}}$. Then we have

$\sum\limits_{n=0}^{\infty} a_{n} (x_a-x_{0})^n= \sum\limits_{n=0}^{\infty} a_{n}(-1)^nr^n$


$\sum\limits_{n=0}^{\infty} a_{n} (x_b-x_{0})^n= \sum\limits_{n=0}^{\infty} a_{n}r^n$.

Then we have:

$\sum\limits_{n=0}^{\infty} a_{n} (x_a-x_{0})^n$ converges absolutely $ \iff \sum\limits_{n=0}^{\infty} a_{n} (x_a-x_{0})^n$ converges absolutely .

  • $\begingroup$ So, are you saying that it is impossible to have this scenario? That is, if one endpoint leads to an absolutely convergent series, then the other endpoint of the interval must also lead to an absolutely convergent series? $\endgroup$ Commented Nov 28, 2019 at 12:12
  • $\begingroup$ Is it even ever possible for a power series to converge absolutely at the end point (on the boundary) of its domain of convergence? I think not. I think the region of absolute convergence is always open. $\endgroup$ Commented Nov 28, 2019 at 12:14
  • $\begingroup$ No. The region of absolute convergence is not always open. Example $\sum_{n=0}^{\infty} \frac{x^n}{n^2}$ This power series converges absolutely in each $x \in [-1,1]$ $\endgroup$
    – Fred
    Commented Nov 28, 2019 at 12:17
  • 1
    $\begingroup$ @MaxFreiburghaus I think it's possible to have such a power series that is absolutely convergent at the endpoints of it's interval of convergence. Consider the following series $\sum\limits_{n=0}^{\infty} \frac{(x-1)^n}{3^n n^2}$, x is in the interval $[-2,4]$. And the series is absolutely convergent at both endpoints of the interval. $\endgroup$ Commented Nov 28, 2019 at 12:21
  • $\begingroup$ @GramCracker: Ah, you're right. Good example. $\endgroup$ Commented Nov 28, 2019 at 12:24

$\sum |a_n| |x_a-x_0|^{n}$ is same as $\sum |a_n| |x_b-x_0|^{n}$.


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