Is this statement regarding power series and limit superior false?

Here is a statement:

Consider the series $\sum a_n x^n$. Suppose $\limsup_{n \to \infty} |\frac{a_n}{a_{n+1}} | = R$, where R is not equal to 0. Then the series converges for all $x \in (-R, R)$.

I believe this statement is false. To make it true, change "limit superior" to "limit inferior".

As a counter-example to the statement as written, consider: $a_n = {1, 1, 3, 3, 3^2 , 3^2 , 3^3 , 3^3 , ...}$ . Then $\limsup_{n \to \infty} |\frac{a_n}{a_{n+1}} | = 1$, but the radius of convergence of the series $\sum a_n x^n$ is $\frac{1}{\sqrt3}$ (By Cauchy-Hadamard).

Am I correct that this original statement is false? Does this counter-example suffice to prove that the original statement is false?

• Your counter example is correct, the statement is false. The limit inferior should work (it will follow $a_n \lessapprox \frac{c}{R^n}$ for large $n$), but will not necessarily give you the radius of convergence. – s.harp Apr 5 '17 at 21:07
• Did you happen to check the details of my counter-example? A professor is claiming the original statement is true. – frito_mosquito Apr 14 '17 at 2:49
• The counter example is very straightforward. $\frac{a_n}{a_{n+1}}=\begin{cases}1&n\text{ odd}\\\frac13&n\text{ even}\end{cases}$ so the limsup is $1$. However if you put in for example $x=\frac1{\sqrt3}$ you get the sum $$\sum_n 3^{n} x^{2n}+\sum_n 3^n x^{2n+1}=\sum_n 1+\sum_n \frac1{\sqrt3}$$ which has no chance to converge absolutely – s.harp Apr 16 '17 at 11:57
• Thank you for taking the time to write that out. This is exactly what I thought as well, but was second guessing myself because the professor disagrees. – frito_mosquito Apr 17 '17 at 18:02