# Question regarding radius of convergence

Let $\sum c_{n}z^n$ be a power series. Now radius of convergence can be found by ratio test and root test. But in ratio test we consider $\lim \dfrac {c_{n}}{c _{n+1}}$. But what will we consider if there are two limit that is if

$\limsup \dfrac {c_{n}}{c _{n+1}}\neq\liminf \dfrac {c_{n}}{c _{n+1}}$. Should we consider lim sup or lim inf.

Consider the series$$1+2z+z^2+2z^3+z^4+2z^5+\cdots$$Its radius of convergence is $1$. On the other hand, if$$c_n=\begin{cases}1&\text{ if n is even}\\2&\text{ if n is odd,}\end{cases}$$then $\limsup_n\frac{c_n}{c_{n+1}}=2$ and $\liminf_n\frac{c_n}{c_{n+1}}=\frac12$. Therefore, the answer is “neither”.
• @user426700 Yes, we can. The radius of convergence is always the inverse of $\limsup_n\sqrt[n]{|c_n|}$. However, in my concrete example, the sequence $\sqrt[n]{|c_n|}$ converges to $1$. Jul 20, 2017 at 9:31
Usually $\limsup |\frac{c_n}{c_{n+1}}| \ge \liminf |\frac{c_n}{c_{n+1}}|$, the equality holds iff the sequence $\left(\frac{c_n}{c_{n+1}}\right)_{n \in \mathbb{N}}$ admit finite limit.
For D'Alembert theorem, to compute the radius of convergence you have to use $\limsup$. If you use $\liminf$ you'll have a radius $r<\rho$ in which the series converge but it is not the radius on convergence $\rho$.