The series in question is:$\frac{1}{2}+1+\frac{1}{8}+\frac{1}{4}+\frac{1}{32}+\frac{1}{16}+\frac{1}{128}+\frac{1}{64}...$
where
$\liminf\limits_{n\rightarrow \infty} \frac{a_{n+1}}{a_n}=\frac{1}{8}$ $\limsup\limits_{n\rightarrow \infty} \frac{a_{n+1}}{a_n}=2$ but $\lim \sqrt[n]a_n=\frac{1}{2}$
I think the series is as follows: $a_{2n-1}=\frac{1}{2^n}$ and $a_{2n}=\frac{1}{2^{n-2}}$.
My idea was to say that suppose $k=2n-1$, an odd number. Then $k+1=2n$ and $\frac{\frac{1}{2^{n-2}}}{\frac{1}{2^n}}=\frac{1}{4}$
I'm not sure what I'm messing up in trying to fill in the details for this example.