Examples of sequences of positive terms $\{a_n\}$ such that $a_n^{1/n}\rightarrow l ~~\text{does not imply}~~ \frac{a_{n+1}}{a_n}\rightarrow l$ 
Give some examples of sequences of positive terms $\{a_n\}$ such that $$a_n^{1/n}\rightarrow l ~~\text{does not imply}~~  \frac{a_{n+1}}{a_n}\rightarrow l$$

If $a_n>0$ for all $n\in \mathbb{N}$, it can be shown that  $\frac{a_{n+1}}{a_n}\rightarrow l $ implies $a_n^{1/n}\rightarrow l$ by using log, a continuous function. This is Cauchy's second limit theorem. But I don't know how to show that its converse (that is the question I have written above) is not true. 
Please help.
 A: Let $\displaystyle a_n=3^{n+(-1)^n}$, so $\{a_n\}=\{3^0, 3^3, 3^2, 3^5, 3^4, 3^7, 3^6, \cdots\}$.
Then $\displaystyle\lim_{n\to\infty}(a_n)^{\frac{1}{n}}=\lim_{n\to\infty}3^{1+\frac{(-1)^n}{n}}=3$,
but $\displaystyle\lim_{n\to\infty}\frac{a_{n+1}}{a_n}$ does not exist, since $\displaystyle\frac{a_{n+1}}{a_n}=\begin{cases}3^3&\mbox {, if n is odd}\\3^{-1}&\mbox{, if n is even}\end{cases}$
A: A simple example is $$ 1, 1, 2, 2, 4, 4, 8, 8, 16, 16, \ldots $$
or more formally $a_n = 2^{\lfloor n/2\rfloor}$.
Then $(a_n)^{1/n} \to \sqrt 2$, but $\dfrac{a_{n+1}}{a_n}$ alternates between $1$ and $2$, and therefore does not have a limit.
A: I am pretty sure the following sequence works:
$$a_1=l$$
$$a_n=\begin{cases}a_{n-1}*(l-1) \ \text{iff} \ n \equiv 0 \pmod 2 \\ a_{n-1}*\frac{l^2}{l-1} \ \text{iff} \ n \equiv 1 \pmod 2\end{cases}$$
Here, what I'm doing is alternating between a ratio of $l-1$ and $\frac{l^2}{l-1}$ so that after every $2$ terms, the ratio is $l^2$, but in between terms, the ratio alternates between $l-1$ and $\frac{l^2}{l-1}$. I am fairly certain this works for $l > 1$, but does not work once $l-1$ is negative because that contradicts that $\{a_n\}$ is positive.
