If $\liminf z_n = 0$, then there is a null sequence $(y_n)$ such that $\sum y_n = \infty$ and $\liminf y_n/z_n = 0$ 
Let $(z_n)$ be a sequence in $(0, \infty)$ with $\liminf z_n = 0$. Show that there are null sequence $(y_n)$ in $(0,\infty)$ such that $\sum y_n = \infty$ and $\liminf y_n/z_n = 0$.

There is a hint told me to construct $(y_n)$ as follows: choose a convergent subsequence $(z_{n_k}) \to 0$. Set
$$
y_{n_k} = z_{n_k}^2 \text{ for all } k \in \mathbb{N}, \quad y_n = 1/n \text{ otherwise}
$$
$(y_k)$ clearly is null and safisfies $\liminf y_n/z_n = 0$. My question is, is $(z_{n_k}) \to 0$ enough to conclude $\sum y_n = \infty$? Since there are instances that can make the sequence converges under this construction (e.g. $z_k = k^{-2}$ and pick subsequence $(z_{k})$ where $k \neq 2^n$ for any $n$), how do we ensure the existence?

What one can do is slighty change the definition of $y_n$. First consider a subsequence of $(z_n)$ convergent to zero. Let $(z_{n_k})$ be that subsequence. Now consider the set of indices $A=\{n_{2k} \mid  k >0\}$ and define the complementary set of indices $B=\{n>0\mid n\notin A\}$. $B$ is an infinite set, since $\{n_{2k-1} \mid  k >0\} \subseteq B$. Now set:
$$y_n=\begin{cases} z_n^2 & n \in A\\\dfrac{1}{\operatorname{card}(\{1,2,\dots n\}\cap B)} & n\in B\end{cases} $$
