Constructing a divergent series Please would you help me with this question? I've been thinking about it for ages but I've made very little headway, so if possible a hint would be ideal. 

Let  $\sum_{n=1}^∞{x_n}$ be a divergent series, where $x_n &gt; 0$ for all $n$. Show that there is a divergent series  $\sum_{n=1}^∞{y_n}$ with $y_n &gt; 0$ for all $n$, such that $(\frac{y_n}{x_n}) → 0.$

I have not been taught analysis formally, hence my lack of progress. I know to consider the series as a sequence of partial sums, and I tried to take the contrapositive of the statement but that just overcomplicated matters. I know I don't have many ideas to present but I have been trying this for days.

Thank you in advance.
You can break up your divergent series $\sum x_n$ into bunches of
consecutive terms, each of which are $&gt;1$. In detail there are
$n_1$, $n_2,\ldots$ with
$$x_1+x_2+\cdots +x_{n_1}&gt;1,$$
$$x_{n_1+1}+x_{n_1+2}+\cdots +x_{n_2}&gt;1,$$
etc. Pick suitable constants $u_1,u_2,\ldots$ and set $y_k=u_ix_k$
whenever $n_{i-1}&lt;k\le n_i$.
To make it even easier:

Define $(y_{n})$ as $y_{1}=\frac{x_{1}}{1}, y_{2}=y_{3}=\frac{x_{2}}{2}, y_{4}=y_{5}=y_{6}=\frac{x_{3}}{3},y_{7}=y_{8}=y_{9}=y_{10}=\frac{x_{4}}{4}\dots$ where every time $\frac{x_{n}}{n}$ $n$ is repeated $n$ times...