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.


2 Answers 2


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”.

  • $\begingroup$ Then we should use root test. But this example actually proves we can not trust on ratio test?? $\endgroup$
    – user426700
    Jul 20, 2017 at 9:19
  • $\begingroup$ @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 $1$. $\endgroup$ 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 $\rho$.


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .