Given a divergent series $\sum x_n$ with $x_{n} \rightarrow 0$, show there exists a divergent series $\sum y_{n} $ with $ y_n/x_n\rightarrow 0$ The problem is what the title says, with the added requrement that both series should have positive terms.
I ruled out defining $y_{n}$ by $x_{n}$ divided by some function of $n$ since I can't see how we can guarantee it's sum is divergent. 
So instead I was looking at defining $y_{n} = max \{ x_{2^{n}} , x_{2^{n}+1} , ... , \} $ but even if this were to work, I have no idea how to prove it.
So can a simple/general $y_{n}$ be found that satisfies these conditions?

Slightly more general result: If $\sum x_n$ is a positive divergent series, then there exists a positive divergent series $\sum y_n$ such that $y_n/x_n \to 0.$ (Note that if in addition we have $x_n$ bounded, then $y_n \to 0.$)
Proof: There are integers $1=n_1 < n_2 < \cdots \to \infty$ such that
$$\sum_{n=n_k}^{n_{k+1}-1} x_n > 1$$
for all $k.$ Let $B_k = \{n: n_k \le n < n_{k+1}\}.$ Define the sequence $y_n$ block by block: If $n\in B_k,$ set $y_n = x_n/k.$ Then
$$\sum_{n=1}^{\infty}y_n = \sum_{k=1}^{\infty}\sum_{n\in B_k}\frac{x_n}{k} = \sum_{k=1}^{\infty}\frac{1}{k}\sum_{n\in B_k}x_n \ge \sum_{k=1}^{\infty}\frac{1}{k}\cdot 1 = \infty.$$
For $n\in B_k$ we have $y_n/x_n = 1/k.$ As $n\to \infty,$ $n$ moves through all blocks $B_1, B_2, \dots$ forcing $y_n/x_n \to 0.$
