Give an example of $\liminf_{k \to \infty} \frac{a_{k+1}}{a_k} \lt \liminf_{k \to \infty} (a_k)^{\frac{1}{k}}$

Question :

Give an example of a sequence $a_k$ with positive numbers such that :
$\liminf_{k \to \infty} \frac{a_{k+1}}{a_k} \lt \liminf_{k \to \infty} (a_k)^{\frac{1}{k}} \lt \limsup_{k \to \infty} (a_k)^{\frac{1}{k}} \lt \limsup_{k \to \infty} \frac{a_{k+1}}{a_k}$

Note 1: The middle inequality is easy but with those two ( right and left ) it seems hard to find an example. Any hint would be great !

Note 2: All the four values of inequality should be finite.

If you want all four numbers to be finite, the following idea can be used instead. Define $$b_k=\begin{cases} 2,&\text{if 2^{n-1}<k\le2^n for odd n},\\ 1,&\text{if 2^{n-1}<k\le2^n for even n},\\ \end{cases}$$ where $k=1$, $2$, ... (so, the sequence of $b_k$ is $1$, $2$, $1$, $1$, $2$, $2$, $2$, $2$ etc.). After that define $$a_k=\prod_{n=1}^kb_n.$$ This way, $\liminf_{k\to\infty}a_{k+1}/a_k=\liminf_{k\to\infty}b_k=1$, $\limsup_{k\to\infty}a_{k+1}/a_k=\limsup_{k\to\infty}b_k=2$.
As for the upper and lower limits of the $n$-th root of $a_n$: $$\liminf_{k\to\infty}a_k^{1/k} =\lim_{n\to\infty}a_{2^{2n}}^{1/2^{2n}} =\lim_{n\to\infty}({2^{1+4+\dots+2^{2n-2}}})^{1/2^{2^n}} =\lim_{n\to\infty}(2^{(2^{2n}-1)/3})^{1/2^{2^n}}=2^{1/3},$$ $$\limsup_{k\to\infty}a_k^{1/k} =\lim_{n\to\infty}a_{2^{2n+1}}^{1/2^{2n+1}} =\lim_{n\to\infty}({2^{1+4+\dots+2^{2n}}})^{1/2^{2^n+1}} =\lim_{n\to\infty}(2^{(2^{2n+2}-1)/3})^{1/2^{2^n+1}}=2^{2/3}.$$ So, the example is a bit more complicated.
If you define $$a_k=\begin{cases} 2^k&\text{if k is odd},\\ 4^k&\text{if k is even}, \end{cases}$$ then \begin{align*} \liminf_{k\to\infty}\frac{a_{k+1}}{a_k}&=\lim_{k\to\infty}\frac{2^{k+1}}{4^k}=0,\\ \liminf_{k\to\infty}a_k^{1/k}&=2,\\ \limsup_{k\to\infty}a_k^{1/k}&=4,\\ \limsup_{k\to\infty}\frac{a_{k+1}}{a_k}&=\lim_{k\to\infty}\frac{4^{k+1}}{2^k}=+\infty. \end{align*}