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 $L = \lim \sup \bigg|\dfrac{c_{n+1}}{c_n}\bigg|$?
[Math] Radius of convergence and ratio test
complex-analysis
Best Answer
Consider the series $$1+2z+3^2z^2+2^3z^3+3^4z^4+2^5z^5+3^6z^6+\cdots.$$ The required limsup of the $n$-th roots is $3$, and the radius of convergence is $\frac{1}{3}$.
Now look at the ratios $\dfrac{c_{n}}{c_{n+1}}$. You will notice they are very badly behaved, and tell us essentially nothing about the radius of convergence.