When I apply the root test to power series I know that

$$R= \frac{1}{\limsup_{n\rightarrow\infty}{\sqrt[n]{|c_n|}}}$$

So if I get $1$ as the result of the limit the radius of convergence of the power series is $R=1$?

Same question also for the ratio test.

I'm confused because when I study the behaviour of numerical series with these criteria and I get 1 as a result it means that they are not applicable.

The prove of the ratio test given on Wikipedia though seems to me clear in indicating that in the power series case $1$ is not a problematic result.

  • $\begingroup$ Yes and yes.${}{}{}{}{}$ $\endgroup$
    – DonAntonio
    Feb 9, 2018 at 11:45
  • $\begingroup$ Yeeeeees don't confused the Radius of convergence and the criteria for simple series $\endgroup$
    – Guy Fsone
    Feb 9, 2018 at 11:51
  • $\begingroup$ Thank you I'll keep in mind $\endgroup$
    – Gitana
    Feb 9, 2018 at 16:07

1 Answer 1


If you apply the ratio test to $\sum c_n z^n$: $$ (\limsup_{n\rightarrow\infty}\root n\of{|c_n|})|z| = \limsup_{n\rightarrow\infty}\root n\of{|c_n z^n|} < 1 $$ implies that the series is convergent for $$|z| < \frac{1}{\limsup_{n\rightarrow\infty}{\sqrt[n]{|c_n|}}}.$$


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