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.
Thanks in advance.
If you want all four numbers to be finite, the following idea can be used instead. Define
$$
b_k=\begin{cases}
2,&\text{if $2^{n-1} As for the upper and lower limits of the $n$-th root of $a_n$:
$$
\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*}