TITLE: Radius of convergence and ratio test
QUESTION [10 upvotes]: My book says that given a power series $\sum_{n = 1}^\infty c_nz^n$ where the $c_n$ are complex the radius of convergence of the series is $\dfrac{1}{L}$ where $L = \lim \sup \sqrt[n]{|c_n|}$.  So the radius of convergence is defined using the root test.  Since we can also apply the ratio test, is it fair to say, that the radius of convergence is $\dfrac{1}{L}$ where $L = \lim \sup \bigg|\dfrac{c_{n+1}}{c_n}\bigg|$?

REPLY [5 votes]: The ratio test is strictly weaker than the root test in the sense that if the ratio test gives an answer, then so does the root test and they are the same.  However, as the other answers show, there are many series for which it gives no answer.  Even though you are using $\limsup$ rather than $\lim$, the limit ratio may fail to be defined if the series contains infinitely many zeros, which make the corresponding ratios undefined, as in julien's answer.
In general, the ratio test is easier for common series because they involve algebraic operations for which computing the $n$th roots is difficult but for which the ratios can be computed pretty easily.  But for theoretical purposes it is inferior.